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NEW  AND  COMPLETE 

SYSTEM  OF  ARITHMETICK. 

COMPOSED 

FOR  THE  USE  OF  THE  CITIZENS 

bF  THE 

UNITED  STATES. 


BY  NICOLAS  PIKE,  A.  M.  A.  A.  S. 


atJID  MlfNtTS   REIPUBLIC^E    MAJUS    MEI-IUSVE   AFFERRE   POSSITMUS,  aiTAM   SI 

JUVENTUTEM  DOCEMUS,  ET  BENE  EUXTDIMUS  ? 

E    VARUS    SUMENDUM    EST    OPTIMUM CICERO. 


FOURTH  EDITION; 
REVISED,    CORRECTED,    AND    IMPROVED, 

BY  CHESTER  DEWEY,  A.A.S. 

PROFESSOR    OF    MATHEMATICKS,  AND   NATURAL   PHILOSOPHY    IW 
WILLIAMS    COLLEGE. 


TROY,  N.  y. 
PRINTED  AND  PUBLISHED  BY  WM.  S.  PAP.KER, 

S07D    AT    THE   TI?OV    BOOKSTORE,  AND  BY  THE  PRINCIPAL    BOOKSELLET?  5    !W 
THE   UNITED    STATES. 


EDUCATION  DEPT. 


.VORTHB^RM'  DISTRICT  OF  XfEW  TORS',  TO  WIT: 

BE  IT  REMEMBERED,  That  on  the  twenty  first  day  of  September,  in  the 
forty  seventh  year  of  the  Independence  of  the  United  States  of  America,  A.  D, 
1822,  William  S.  Parker  of  the  said  District,  hrts  deposited  in  this  Office  the 
title  of  a  Book,  the  right  whereof  he  claims  as  proprietor,  in  the  words  follow- 
ing', to  wit : 

"  A  new  and  comple  system  of  Arithmetick,  composed  for  the  use  of  the  citi  • 
zens  of  the  United  States.  By  Nicholas  Pike,  a.  m.  a.  a.  s.  Quid  munus 
reipublicffi  majus  meliusve  afferre  possumus,  quam  si  juventutem  docemus,  et 
bene  erudimus  ?  E  variis  sumendum  est  optimum. — Cicero.  Fourth  Edition ; 
revised,  corrected,  and  improved,  by  Chester  Dev»^et,  a.  a.  s.  Professor  of 
Mathematicks,  and  Natural  Philosophy  in  Williams  College." 

In  conformity  to  the  act  of  the  Congi-ess  of  the  United  States,  entitled  "  An 
act  for  the  encouragement  of  learning,  by  securing  the  copies  of  Maps,  Charts, 
and  Books,  to  the  authors  and  proprietors  of  such  copies,  during  the  times  there- 
in menfioned  ;"  and  also,  to  the  act  entitled  "  An  act  supplementary  to  an  act 
/mtitled  '  An  act  for  the  encourag-ement  of  learning,  by  securing  the  copies  of 
Maps,  Charts,  and  Books,  to  the  authors  and  proprietors  of  such  copies  during 
the  times  thereiii  mentioned,'  and  extending  the ,  bQuefits  thereof  to  the  arts  ot 
Designing,  Engraving  anlCt^tltjng  hisXotJcal.aqd;ot|ier  prints." 

'  v^         .RICHMD'.R.  LANSING, 

CUx^i  qf  J^LG  Northern  District  of  Js'ew  YorJi- 


PREFACE 

TO  THE  FIRST  EDITION. 

It  maji,  perhaps,  by  some,  be  thought  needles?,  when  Au- 
thors are  so  muhiphed,  to  attempt  pubhshing  any  thina^  further  on 
Arithmetick,  as  it  may  be  imagined  (here  can  be  nothing  more  than 
the  repetition  of  a  subject  already  exhdusted.  It  is  however  the 
opinion  of  not  a  few,  who  are  conspicuous  for  their  knowledge  in 
the  Mathematicks,  that  the  books,  now  in  use  among  us,  are  gen- 
erally deficient  in  the  illustration  and  application  of  the  rules  ;  of 
the  truth  of  which,  the  general  complaint  among  Schoolmasters  is  a 
strong  confirmation.  And  not  only  so,  but  as  the  United  States  are 
now  an  independent  nation,  it  was  judged  that  a  System  might  b,e 
calculated  more  suitable  to  our  meridian,  than  those  heretofore 
published. 

Although  I  had  sufficient  reason  to  distrust  my  abilities  for  so  ar- 
duous a  task,  yet  not  knowing  any  one  who  would  take  upon  him- 
self the  trouble,  and  appreht-nding  I  could  not  render  the  publick 
more  essential  service,  than  by  an  attempt  to  remove  the  difficul- 
ties complained  of,  with  diffidence  1  devoted  myself  to  the  work. 
.  I  have  availed  my?elf  of  the  best  authors  which  could  be  obtained 
but  have  followed  none  particularly,  except  Bonnycastle's  Method 
of  Demonstration. 

Although  I  have  arranged  the  work  in  such  order  as  appeared  to 
me  the  most  regular  and  natural,  the  student  is  not  obliged  to  pay  a 
strict  adherence  to  it ;  but  may  pass  from  one  Rule  to  another,  as 
his  inclination  or  opportunity  for  study,  may  reqiiire. 

The  Fpderal  Coin,  being  purely  decimal,  most  naturally  falls  in 
after  Decimal  Fractions.  t 

I  have  given  several  methods  of  extracting  the  Cube  Root,  and 
am  indebted  to  a  learned  friend,  who  declines  having  his  name 
made  publick,  for  the  investigation  of  two  very  concise  Algebraick 
Theorems  for  the  extraction  of  all  Hoots,  and  of  a  particular  The- 
orem for  the  Sursolid. 

Amnng  the  Miscellaneous  Questions,  I  have  given  some  of  a  phi- 
losophical nature,  as  vv.ll  with  a  vi(;vv  to  inspire  the  pupil  with  9k- 
relish  for  philosophical  studies,  as  to  the  usefulness  of  thew  in  the 
common  business  of  life. 

Being  sensible  the  followirjg  Treatise  will  stand  or  fall,  accord* 
ing  to  its  real  merit  or  demerit,  I  submit  it  to  the  judgment  of  the 
candid- 

With  pleasure  I  embrace  this  opportunity,  to  express  my  grati- 
tude to  those  learned  Gentlemen,  who  have  honoured  tliis  Treatise 
with  their  approbation,  as  well  as  to  such  Gentlemen,  as  ha.e  en- 
couraged it  by  their  subscriptions  ;  and  to  request  the  reader  to 
excuse  any  errours  he  may  meet  with  ;  for  although  great  pains  have 
been  taken  in  correcting,  yet  it  is  difficult  to  prevent  errours  from 
creeping  into  the  press,  and  some  may  have  escaped  my  own  ob 
servation  ;  in  cither  ca*;e,  a  hint  from  ihe  candid  will  much  oblige 
their  Most  obedient, 

And  humble  Servant, 

THE  AUTHOR 


ir>^r%itr%^  ¥%nr% 


PREFACE 

TO  THE  FOfFRTH  EDITION. 

Pike's  Anrr^HMETicK  is  universally  acknowledged  to  be  the  most 
complete  system  ever  published  in  the  United  States.  It  early 
obtained  a  very  high  reputation,  and  has  continued  to  receive  the 
approbation  of  the  publick,  wherever  it  has  been  used.  It  is  de- 
signed for  the  instruction  of  our  youth  in  academies  and  higher 
schools,  as  well  as  for  the  use  of  the  man  of  business  and  the  gen- 
tleman. All  those  rules,  which  are  so  frequently  employed  in  the 
various  departments  of  business,  are  introduced  into  it.  It  is  the 
source  too,  from  which  the  later  Arithmelicks  have  chiefly  been 
compiled.  By  them,  however,  it  has  not  been  superseded,  so 
much  more  full  and  extensive  are  its  rules  and  their  application. 
In  the  demonstration  and  illustration  of  the  rules,  it  stands  pre- 
eminent. 

The  continued  demand  for  the  work  has  induced  the  publisher 
and  proprietor  of  the  copy  right,  to  present  to  the  publick  a  new 
and  improved  edition.  In  the  revision  of  the  work  much  labour 
has  been  bestowed,  and  in  the  language  of  a  Mathematician  well 
acquainted  with  the  work,  "  to  excellent  purpose.  It  is  still  Pike's 
Arithmetick,  but  altogether  more  perfect  than  it  was  before. 
As  a  complete  system,  it  may  be  pronounced  superior  to  any  ever 
published."  The  imperfections  of  the  previous  editions,  which 
have  been  noticed  by  the  most  distinguished  teachers  of  Arithme- 
tick, are  to  a  great  degree  remedied  in  the  present  edition. 

The  alterations  and  improvements  consist  in  the  following  par- 
ticulars. Several  rules  have  been  added,  as  well  as  a  variety  of 
Tables,  of  much  practical  importance.  Some  Tables  have  been 
corrected  and  others  have  been  enlarged.  Several  simple  and 
obvious  rules  were  redundant  and  have  been  omitted.  The  Rule 
of  Three  and  Interest  have  been  much  improved*  .Demonstra- 
tions of  a  large  proportion  of  the  rules  were  riot  givein  by  Mr. 
Pike  :  where  the  subject  would  readily  admit,  they  have  been 
supplied.  The  illustrations  of  the  Kules  are  more  copious,  and 
in  many  cases  simplilied.  Most  of  the  Algebraiok  demonstrations, 
which  are  useless  to  the  mere  student  in  Arithmetick,  have  been 
exchanged  for  arithmetical  illustrations.  I^ogarithms,  TrigonomC" 
try,  Algebra,  and  Conic  Sectiofis,  are  omitted.  These  subjects 
were  so  briefly  treated  by  Mr.  Pike,  as  to  possess  little  value.  As 
they  require  a  large  volume  of  themselves,  and  are  very  fully 
treated  of  in  Day's  Course  of  Alathematicks,  and  in  the  system  of 
Mathematicks  now  publishing  at  the  University  in  Massachusetts, 
the  publisher  has  been  uniformly  advised  to  omit  them  entirely. 

A  concise  System  of  Book  Keeping  by  single  and  double  Entry, 
has  been  added  to  the  work,  which,  we  hesitate  not  to  say,  will 
ginatly  enhance  its  value. 

U  is  confidently  believed  that  this  edition  will  merit  the  appro- 
bation of  the  publick,  and  receive  that  patronage  vyhich  has  been 
so  liberally  bestowed  on  the  previous  editions. 

THE  PUBLISHER. 

Troy,  October  31,  1821:. 


RECOMMENDATIONS. 


Dartmouth  University,  1786. 
At  the  request  of  Nicolas  Pike,  Esq.  we  have  inspected  his  System  of  Arith- 
metiek,  which  we  cheerfully  recommend  to  the  publick,  as  easy,  accurate,  and 
complete.     And  we  apprehend  there  is  no  treatise  of  the  kind  extant,  from 
which  so  great  utility  may  arise  to  Schools. 

B.  WOODWARD,  Math,  and  Phil.  Prof. 
JOHN  SMITH,  Prof,  of  the  Learned  Languagea. 

I  do  most  sincerely  concur  in  the  preceding  recommendation. 

J.  WHEEJ^OCK,  President  of  the  University. 


Providence,  Rhode  Island,  1785. 

Whoever  may  have  the  perusal  of  this  treatise  on  Ai'ithmetick  may  natur- 
ally conclude  I  might  have  spared  myself  the  trouble  of  giving  it  this  recom- 
mendation, as  the  work  will  speak  more  for  itself  than  the  most  elaborate  rec- 
ommendation from  my  pen  can  speak  for  it :  But  as  I  have  always  been  much 
delighted  with  the  coutempljftion  of  mathematical  subjects,  and  at  the  same 
time  fully  sensible  of  the  utility  of  a  work  of  this  nature,  1  was  willing  to  render 
every  assistance  in  my  power  to  bring  it  to  the  publick  view  :  And  should  the 
student  read  it  with  the  same  pleasure  with  which  I  perused  the  sheet?  before 
they  went  to  the  press,  I  am  persuaded  he  will  not  fail  of  reaping  that  benefit  from 
it  which  he  may  expect,  or  wish  for,  to  satisfy  his  curiosity  in  a  subject  of  this 
nature.  The  author,  in  treating  on  numbers,  has  done  it  with  so  much  perspicu- 
ity and  singular  address,  that  I  am  convinced  the  study  thereof  will  become 
more  a  pleasure  than  a  task. 

The  arrangement  of  the  work,  and  the  method  by  which  he  leads  the  fi/ro 
into  the  first  principles  of  xiumbcrs,  are  novelties  I  have  not  met  with  in  any  book 
i  have  seen.  Wingate,  Hutton,  Ward,  Hill,  and  many  other  authors  whose 
names  might  be  adduced,  if  necessary,  have  claimed  a  considerable  siiare  of  mer- 
it ;  but  when  brought  into  a  compariktive  point  of  view  with  this  treatise,  they 
are  inadequate  and  defective.  This  volume  contains,  besides  what  is  useful 
and  necessary  in  tlie  common  affairs  of  life,  a  great  fund  for  amusement  and  en- 
tertainment. The  Mechanick  will  find  in  it  mvich  more  than  he  may  have  oc- 
casion for  ;  the  Lawyer,  Merchant  and  Mathematician,  will  find  an  ample  field 
for  the  exercise  of  their  genius  ;  and  I  am  well  assured  it  may  be  read  to  great 
advantage  by  students  of  every  class,  from  the  lowest  school  to  the  University. 
More  than  this  need  not  be  said  by  me,  and  to  have  said  less,  would  be  keeping- 
back  a  tribute  justly  due  to  the  merit  lof  this  work. 

BENJAMIN  WEST. 


University  in  Cambridge,  1786. 
Having,  by  the  desire  of  Nicolas  Pike,  Esq.  inspected  the  following  volume 
in  manuscript,  we  beg  leave  to  acquaint  the  publick,  that  in  our  opinion  it  is  a 
work  well  executed,  and  contains  a  complete  system  of  Arithmetick.  The 
rules  are  plain,  and  the  demonstrations  perspicuous  and  satisfactory  ;  and  we  es- 
teem it  the  best  calculated,  of  any  single  piece  we  have  met  with,  to  lead  youth, 
by  natural  aad  easy  gradations,  into  a  methodical  and  thorough  acquaintance 
with  the  science  of  figures.  Persons  of  all  descriptions  may  find  in  it  every  thing, 
respecting  numbers,  necessary  to  their  business;  and  not  only  so,  but  if  they 
have  a  speculative  turn,  and  mathematical  taste,  may  meet  v/itb  much  for  their 
entertciinmeat  at  a  leisure  hour. 


vi  RECOMMENDATIOiNS. 

We  are  happy  to  see  so  useful  an  American  production,  which,  if  it  should 
meet  with  the  encouragement  it  deserves,  among  the  inhabitants  of  the  United 
States,  will  save  much  money  in  the  country,  which  would  otherwise  be  sent 
to  Europe,  for  publications  of  this  kind. 

We  heartily  recommend  it  to  schools,  and  to  the  communily  at  large,  and 
wish  that  the  industry  and  skill  of  the  Author  may  be  rewarded,  for  so  benefi- 
cial a  work,  by  meeting  with  the  general  approbation  and  encouragement  of  the 
publick. 

JOSEPH  WILLARD,  D.  D.   President  of  the  University. 

E.  WIGGLESWORTII,  S.  T.  P.  HoUis. 

S.  VVILLTAMS,  L.  L.  D.  Math,  et  Phil.  Nat.  Prof.  HoUis. 


Yale  College,  1786. 
Upon  examining  Mr.  Pike's  System  of  Arithmetick  and  Geometry,  in  manu- 
script, I  find  it  to  be  a  work  of  such  mathematical  ingenuity,  that  I  esteem  myself 
honoured  in  joining  with  the  Rev.  President  Willard,  and  other  learned  gentle- 
men, in  recommending  it  to  the  publick  us  a  production  of  genius,  interspersed 
with  originality  in  this  part  of  learning,  and  as  a  book,  suitable  to  be- taught  ia 
schools  :  of  utility  to  the  merchant,  and  well  adapted  even  for  the  University  in- 
struction, I  consider  it  of  such  merit,  as  that  it  will  probably  gain  a  very  gene- 
ral reception  and  use  throughout  the  republick  of  letters. 

EZRA  STILES,  President. 


BosTOBT,  1786. 
From  the  known  character  of  the  Gentlemen  who  have  recommended  Mr. 
Pike's  System  of  Arithmetick,  there  can  be  no  room  to  doubt,  that  it  is  a  valua- 
ble performance  ;  and  will  be,  if  published,  a  very  useful  one.     I  therefore  wish 
him  success  in  its  publication. 

JAMES  BOWDOIN. 


Union  College,  Oct.  10, 1822. 
Pike's  Arithmetick  is  too  well  known  and  too  highly  appreciated  to  re? 
quire  any  recommendation  ;  and  by  furnishing  an  edition  of  that  work,  in  which 
oommon  language  is  :;ubstituled  for  algebraic  signs,  Professor  Dewey  has  confer- 
red a  favcar  on  those  who  may  wish  to  acquire  or  teach  Aritlimetick  witho  i:f 
jUgebra ;  by  whom  it  is  presumed  this  edition  will  be  patronised. 

E.  NOTT,  President. 


Schenectady,  Oct.  16,  1822. 
Mr.  Wm.  S.  Parker, 
I  have  for  many  years  been  fully  acquainted  with  P^7rc'5  Si/stem  ofArithme- 
tick,  and  am  persuaded  of  its  excellence  ;  I  do  not  know  of  any  treatise  of  more 
practical  utility  ;  the  arrangements  of  its  parts  is  natural,  its  rules  are  plain  and 
easily  understood  and  applied,  and  it  contains  all  that  is  ofany  importance  to  the 
Mercantile  or  Scientific  Arithmetician.  •  To  those  who  have  not  the  elementary 
knowledge  of  Algebra,  the  translation  of  the  Algebraic  expression  into  plain 
Arithmetical  language  must  be  very  acceptable  and  profitable.  This  improve- 
ment, together  with  the  notes  and  emendations  of  Professor  Dewey,  cannot  fail 
to  ensure  the  public  confidence  and  patronage,  A  hand  so  able  as  his,  cannot 
touch  without  improving  an  elementary  treatise,  and  wherever  he  is  known,  his 
«'ime  must  be  a  sufficient  credential. 

Wishing  you  all  success,  and  abundant  remuneration  for  your  labours,  I  am, 
Sir,  your  friend  and  servant. 

T.  M'AULEY,  S.  T.  D. 
Late  Professor  of  Mathematicks.  Union  College. 


RECOMMENDATIONS.  vii 

Amherst,  Mass,  Feb.  9, 1822. 
I  HAVE  long  been  acquainted  with  Pike's  Arithmetic,  and  think  it  the  best 
of  any  extant,  for  those  who  wish  to  acquire  a  thorough  knowledge  of  Arith- 
metic as  a  science  and  an  art.  The  plan  of  improvement  adopted  and  pursued 
by  Professor  Dewey,  in  the  present  edition,  is,  in  my  opinion,  such  as  to  render 
the  work  more  perfect  and  more  useful.  By  supplying  defects,  omitting  redun- 
dancies, and  illustrating  what  was  obscure,  he  has  given  to  the  present  edition  u 
superior  value.  I  cheerfully  recommend  the  work  to  the  patronage  of  the  pub- 
He,  and  especially  to  the  patronage  of  the  Instructors  of  youth  in  Academies  and 
Siohools,  as  combining  more  excellences  than  any  other  Arithmetic  now  hi  use. 

ZEPH.  SWIFT  MOORE, 
President  of  the  Collegiate  Institution,  at  Amherst,  Mass. 


Lewox,  Ms.  April  20,  1822. 
I  HAVE  seen  Pike's  Arithmetic  revised  by  Mr.  Professor  Dewey  of  Williams 
College.  I  entirely  approve  of  all  the  alterations,  fidditions,  and  jllustrations. 
I  cannot  but  believe,  that  the  work  thus  presented  to  the  public,  will  be  duperi- 
or  to  any  thing  of  the  kind  extant.  While  it  initiates  the  scholar  into  the 
theory  of  this  science,  it  is  distinguished  for  a  happy  conciseness,  lucid  method, 
and  graceful  -simplicity,  which  cannot  fail  to  make  it  a  valuable  companion  for 
rtbe  Merchant,  Mechanic,  or  Fanner. 

LEVI  GLEZEN, 
Preceptor  of  Lenox  Academy. 


Extract   of  a   Letter  from  Mr.  Benedict,  Tutor  of  JVilliams  College,  to  the 
Publisher,  dated 

Williams  College,  Janitary  2,  1822. 
Mr.  Parker, 

"  From  the  experience  which  I  have  had  in  instructing  youth,  I  have  had  occa- 
sion to  acquaint  myself  with  many,  if  not  most  of  the  Systems  of  Arithmetick  in 
use  in  this  country.  I  can  therefore  speak  with  some  more  coniidence  than  I 
otherwise  should,  iVom  having  proved  their  excellences  and  defects  by  actual 
trial  of  them.  It  is  most  certain  that  as  a  complete  System  on  this  important 
part  of  education,  the  work  under  consideration  stands  preeminent.  It  is  impos- 
sible that  Arithmetick  should  be  so  treated  ol",  as  not  to  leave  much  to  be  done 
by  the  instructor.  Still,  as  I  think.  Pike's  System  will  enable  the  teacher  to  ben- 
efit his  scholars,  to  give  them  sound  theoretical  and  practical  knowledge  in  this 
branch,  to  induce  them  to  think  and  reason  closely,  and  increase  their  power  of 
arithmetical  invention,  far  more  than  any  one  within  the  compass  of  my  knowl- 
edge. Excellent  as  it  was  when  it  came  from  its  author,  it  had  its  defects.  By 
the  revfsion  of  it  by  Lord,  little  else  was  done  than  to  change  the  sterling  to  fed- 
eral notation.  Much  remained  to  be  done.  In  some  parts,  Mr.  Pike  had  been 
needlessly  minute,  and  loaded  the  work  with  a  multiplicity  of  rules  on  one  sub- 
ject, which  the  accountant  could  not  but  make  for  himself,  as  occasion  demanded, 
with  perfect  ease.  Though  his  illustrations  and  demonstrations  are  usually  very 
good,  in  some  cases  tliey  were  obscure  ;  and  in  some  parts,  as  for  instance  that  of 
interest,  there  was  a  great  deficiency.  I  have  examined  the  work  with  Mr. 
Dewey's  corrections,  with  considerable  care.  He  has  bestowed  great  labour 
upon  it,  and  I  think  to  excellent  purpose.  It  is  still  Pike's  Arithmetick  ;  but 
altogether  more  perfect  than  it  was  before.  I  do  beheve  that  as  a  complete  Sys- 
tem, it  may  be  pronounced  superior  to  any  one  ever  published.  I  most  earnestly 
wish  you  success  in  its  publication  ;  and  I  feel  a  confidence  that  good  judges  will 
not  hesitate  on  perusing  it,  tog;ive  it  an  unqualified  recommendation." 

Yours  respectfully, 

GEORGE  W.  RKXEDTCT. 


CONTENTS. 


Page. 

NUMERATION                     ....  ^^ 

Simple  Addition                        u                   .  .  -  20 

, .    Subtraction                    -                    .  ,  .  ^4 

Multiplication               .                   -  .  .  ^5 

Division                         -                   -  .  .  22 

Tables  in  Compound  Addition                    _  *  -  ^^ 

Compound  Addition                 -                   -  -  "50 

—     Subtraction            -                   -  -  "  54 

Problems  resulting  from  the  preceding  Rules  -  -  ^^ 

Reduction               -                   -                   -  -  "60 

Vulgar  Fractions                       -                   -  -  -  7^ 

Decimal  Fractions                    ....  ^ 

Decimal  Tables                         -                   -  -  -  \q\ 

Compound  Multiplication                           ^  .  .  -jq^ 

■     Division                                     -  -               .    -'  joS 

Reduction  of  Coins                   -                   -  -  -  114 

Duodecimals,  or  Cross  Multiplication        -  -  -  126 

Single  Rule  of  Three                -                   -  -  -  129 

The  method  of  making  Taxes,  in  a  Note  -  -  136 

Single  Rule  of  Three  Direct  in  Vulgar  Fractions        -  -  137 

To  find  the  value  of  Gold  in  New  England  currency  -  -  141 

Single  Rule  of  Three  Direct  and  Inverse  -  -  144 

Rule  of  Three  Inverse                                 -  -  -145 

Abbreviation  of  Vulgar  Fractions              -  -  -  147 

Double  Rule  of  Three                                 -  -  -  148 

Conjoined  Proportion               -                   -  -  -  152 

Arbitration  of  Exchanges        -                   -  -  .  154 

Single  Fellowship                     ....  155 

Double  Fellowship                   -                   -  -  -  139 

Practice                  -----  161 

Form  of  a  Bill  of  Parcels         -                   -  -  -  184 

Tare  and  Tret                           -                   -  -  -  184 

Involution               -                   -                   -  -  -  186 

Evolution               -                   -                   -  -  -  188 

Table  of  Powers                     -                   -  -  -  189 

Extraction  of  the  Square  Root                  -  -  -  190 

Application  and  Use  of  the  Square  Root  -  -  194 

Extraction  of  the  Cube  Root                    _  _  -  197 

Application  and  use  of  the  Cube  Root     -  -  -  203 

Extraction  of  the  Biquadrate  Root          _  -  -  205 

of  the  Sui-solid  bv  Approximation  -  -  206 

Of  the  Roots  of  all  Powers     _  -  -  507 

Surds                      -                 -                   -  -  -  210 

Proportion  in  General           -                  -  -  -  217 

Arithmetical  Proportion        -                   -  -  -  219 

Progression        _                   _  -  -  220 

Geometrical  Proportion          _                  _.  _  -  233 

Progression        -                   -  -  -  235 

Simple  Interest                        _                  ^  _  -  253 

Commission  or  Factorage       ..                   _  _  -  255 

Brokerage                                -                   _  _  -  256 

Buying  and  Selling  Storks     -                  _  _  -  256 

Simple  Interest  by  Decimals                     _  _  _  261 

in  Federal  Money           _  _  _  257 


CONTENTS.  ix 


Rule  for  Interest  on  Bonds,  &c.  -  *■  -  268 

in  Massachusetts  ^  -  -  -  269 

in  Connecticut    -  -  "  *  -  270 

in  New  York      -----  270 

Discount  .  ,  .  -  -  275 

by  Decilnals  .  -  -  -  279 

Duties  .  »  .  -  _  281 

Barter  _  .  ^  -  -  -  282 

Loss  and  Gain  -  -  -  -  -  284 

Equation  of  Payments  .  -  -  _  2B9 

by  Decimals  -  -  -  291 

Exchange  -  -  -  -  -  293 

Policies  of  Insurance  _  -  -  -'  297 

Compound  Interest  .  .  .  -  3Q4 

by  Decimals  -  -  -  3Q7 

Discount  by  Compound  Interest  -  -  -  312 

Annuities  -  -  -  -  -  313 

Annuities,  or  Pensions  in  Arrears,  at  Compound  Interest  -  31s 

Present  Worth  of  Annuities  at  Compound  Interest  -  320 

Annuities  in  Reversion  at  Compound  Interest  -  -  324 

Purchasing  Annuities  forever,  or  Freehold  Estates  -  330 

Purchasing  Freehold  Estates  in  Reversion  -  -  332 

Table,  shewing  the  amount  of  jGl  or  $1  from  1  year  to  50  -  334 

shewing  the  present  worth  of  £l  or  $1  from  I  year  to  40  336 

shewing  the  amount  of  £1  or  $\  Annuity,  &c.  -  337 

shewing  the  present  worth  of  £1  or$l  Annuity,  &c.  -  338 

shewing  the  Annuity  which  £1  or  $1  will  purchase,  &c.  339 

value  of  an  Annuity  of  £1  or  $\  at  different  rates  and  payable  at 

different  periods  340 
value  of  an  Annuity  of  £1  or  $1  for  a  single  life  -  341 
Expectation  of  Life  at  several  ages                  .                 _  _  342 
Circulating  Decimals                        .                 .                 _                 -  342 
Alligation  Medial            -                 -                 -                 .  -^  348 
Alligation  Alternate                         -                 -                 .                 -  3^49 
Single  Position                 -                  -                 -                 .                  _  354 
Double  Position              -                 -                 -                 .  .  356 
Permutations  und  Combinations                      -                 .                 _  359 
A  short  method  of  reducing  a  Vulgar  Fraction  to  a  Decimal           -  364 
of  finding  the  Duplicate,  Triplicate,  &c.  Ratio  of  two  Num- 
bers, whose  difference  is  small          -  -  oge 
To  estimate  the  Distance  of  Objects                               -  -  ogiy 
To  estimate  the  Height  of  Objects              -                   .  .  ^g 
Miscellaneous  Questions                               ,                   .  ^^o 
Of  Gravity                        -                  .                  .  I^^'' 
Ofthe  Fall  of  Bodies                           .                   -  ,  «77 
Of  Pendulums                   -                   _                   .  _  '^ 
Of  the  Lever  or  Steelyard                  -                   _  _ 
Ofthe  Wheel  and  Axle                       -                   .  .  %o 
Ofthe  Screw                    .... 
Of  the  Specifick  Gravities  of  Bodies 
Table  of  Specifick  Gravities                        -                   -  _  30^ 
Use  ofthe  Barometer  in  Measuring  Heights                -  ^  29"^ 
Legal  value  of  Foreign  Coins                      -                   -  -  394 
Tableoftiic  Weight  of  Money                   -                   .  .  395 
of  Exchange                   -                   -                   _  .  39^5 
ofthe  Value  of  Sundry  Pieces  in  the  several  States  -  390 
of  Commission  or  Brokerage            -                   .  -  399 
.     of  the  net  PrOcoe^ls,  after  the  Commispiolis  at  ^h  and  5  Pf  T  cent  ?.rc 
cleductM 


381 

383 


?84 

386 


B 


JO'^ 


X  CONTENTS. 

Page. 
Table  shewing  the  number  of  Days  from  any  day  in  one  month  to  any  day 

in  any  other  month  401 

Table  of  the  Measure  of  Length  of  the  principal  places  in  Europe  compared 

with  the  American  Yard  402 

Table  directing  how  to  buy  and  sell  by  the  Hundred  Weight  403 

Comparison  of  the  American  Foot  with  the  Feet  of  other  Countries  404 

Table  to  cast  up  Wages  or  expenses  for  a  yr.  at  so  much  per  day,  week,  or  mo.  405 

Table  to  find  Wages,  or  Expenses  for  a  mo.  week  or  day,  at  so  much  per  yr.  405 

A  Perpetual  Almanack  _  _  _  .  4Qg 

Tables  reducing  Troy  Weight  to  Avoirdupois,  and  the  contrary  407 

of  the  Money  of  Commercial  Countries  -  -  408 

Jews,  Greeks,  and  Romans  -  418 

of  Measures  of  Length,  Capacity,  &c.  of  Various  countries  419 

An  Account  of  the  Gregorian  Calendar,  or  New  Style  -  425 

Chronological  Problems  426 

Problem  1.  To  find  in  which  Century  the  last  year  is  to  be  Leap  year,  and 

the  contrary  426 

Prob.  2.  To  find,  with  regard  to  any  other  Years,  whether  any  Year  be  Leap 

Year,  or  not  426 

3.  To  find  the  Dominical  Letter  according  to  the  Julian  Method  426 

4.  To  find  the  same  according  to  the  Gregorian  Method  427 

5.  To  find  the  Prime,  or  Golden  Number  -  428 

6.  To  find  the  Juhan  Epact  -  -  428 

7.  To  find  the  Gregorian  Epact  -  -  429 
To  find  the  same  forever             _                  _                  _  430 

8.  To  calculate  the  Moon's  Age  on  any  given  Day  -  430 

9.  To  find  the  Times  ofthe  New  and  Full  Moon,  and  first  and  last  Qrs.  431 

10.  Having  the  time  of  the  Moon's  Southing  given,  to  find  her  Age        432 

11.  To  find  the  time  ofthe  Moon's  Southing  432 

12.  To  find  on  what  day  of  the  week  any  given  day  in  any  rfio.  will  fall  433 

13.  To  find  the  Cycle  ofthe  Sun  434 
Table  ofthe  Dominical  Letters  according  to  the  Cycle  of  the  Sun  434 
Prob.  14.  To  find  the  year  of  the  Dionysiau  Period                        -  434 

15.  To  find  the  year  of  Indiction                        -  -  435 

16.  To  find  the  Julian  period         -                   _  _  435 

17.  To  find  the  Cycle  of  the  Sun,  Golden  Number  and  Indiction,  &c.       435 

18.  To  find  the  time  of  High  Water                   -  -  436 

19.  To  find  on  Avhat  day  Easter  will  happen  -  -  436 
Table  to  find  Easter  from  the  year  1753  to  4199  -  -  438 
Plane  Geometry  _  ,  _.  _  439 
Mensuration  of  Superficies  and  Solids  _  -  -  443 
Section  I.  Of  Superficies  _  _  _  443 
Article  1 .  To  measure  a  Square               _                   _  _  443 

2.  To  measure  a  Parallelogmm,  or  Long  Square  443 
3-  When  the  Breadth  of  a  Superficies  is  given  to  find  how  much 

in  Length  will  make  a  Square  Foot,  Yard,  &c.  444 

4.  To  measure  a  Rhombus          -                    -                   -  .              444 

5.  To  measure  a  Rhomboides                         -                  -  445 

6.  To  measure  a  Triangle                                 -                   -  445 

7.  Another  method  of  finding  the  Area  of  the  Triangle  447 

8.  To  measure  a'  Trapezium       -                  _                  _  447 

9.  To  measure  any  Irregular  Figure              -                   -  448 
10-  To  measure  a  Trapezoid                            -                  ^  449 

11.  To  measure  any  Regular  Polygon  -  -  449 

12.  Having  the  Diameter  of  a  Circle,  to  find  the  Circumference        451 

13.  Having  the  Circumference,  to  find  the  Diameter  452 

14.  To  find  the  Area  of  a  Circle  45S 

15.  Having  the  Diam.  to  find  the  Aroa,  without  the  Circumference  453 
1 G.  Having  the  Circum.  to  find  the  Area  without  the  Diameter         454 


CONTENTS,  s\ 

Page. 
Article  17.  Having  the  Dimensions  of  any  of  the  parts  of  a  Circle,  to  find 

the  Side  of  a  Square,  equal  to  the  Circle                  -  454, 

18.  Having  the  Area  of  a  Circle,  to  find  the  Diameter     -  455 

19.  Having  the  Area,  to  find  the  Circumference               -•  455 

20.  Having  the  Side  of  a  Square,  to  find  the  Diameter  of  a  Circle, 

which  shall  be  equal  to  the  Square  whose  side  is  given  45a 

21.  Having  the  Side  of  a  Square,  to  find  the  Circumference  of  a  Cir- 

cle equal  to  the  given  Square                                             '  45fi 

22.  Having  the  Diam,  of  a  Circle,  to  find  the  Area  of  a  Semicircle  456 

23.  Having  the  segment  of  a  circle,  to  find  the  length  of  the  Arch  Line  45G 

24.  Having  the  Chord  and  Versed  Sine  of  a  Segment,  to  find  the  Di- 

ameter of  a  Circle  457 

25.  To  measure  a  Sector                                   -                   -  457 

26.  To  measure  the  Segment  of  a  Circle        -                   -  458 

27.  To  measure  an  Ellipsis                               -                   -  460 
Directions  for  applying  Superficies  to  Surveying         -  460 

Section  n.     Of  Solids                                -                  -                   -  461 

Art.  28.  To  measure  a  Cube                     _                   _                   -  4G1 

29.  To  measure  a  Parallelopipedon                     -                  -  463 
Having  the  Side  of  a  Square  Solid,  to  find  what  Length  will  make 

a  Solid  Foot  465 

30.  To  measure  a  Cylinder               -                  -                   -  465 
Having  the  Diameter  of  a  Cylinder  given,  to  find  what  length  will 

make  a  Solid  Foot  466 
f            To  find  how  much  a  round  tree,  which  is  equally  thick,  from  end  to 

end,  will  hew  to  when  made  square           -                   -  46t» 

31.  To  measure  a  Prism                     _                   _                   -  467 
.32.  To  measure  a  Pyramid  or  Cone                      -                  -  467 

33.  To  measure  the  Frustum  of  a  Pyramid  or  Cone               -  469 

34.  To  measure  a  Sphere  or  Globe         _             -                   -  472 

35.  To  measure  a  Frustum  or  Segment  of  a  Globe                 -  473 

36.  To  measure  the  middle  Zone  of  a  Globe                          —  473 

37.  To  measure  a  Spheroid                                   —                   -  474 

38.  To  measure  the  middle  Frustum  of  a  Spheroid               -  474 

39.  To  measure  a  Segment,  or  Frustum  of  a  Spheroid           -  474 

40.  To  measure  a  Parabolick  Conoid                   -                   -  474 

41.  To  measure  the  lower  Frustum  of  a  Parabolick  Conoid  475 

42.  To  measure  a  Parabolick  Spindle  475 

43.  To  measure  the  middle  Zone,  or  Frustum  of  a  Parabolick  Spindle  4T5 
-44.  To  measure  aCylindroidor  Prismoid            -                   —  475 

45.  To  measure  a  solid  Ring             _                   _                  _  476 

46.  To  measure  the  Solidity  of  any  ijiregular  Body  whose  Dimensions 

cannot  be  taken                    _                  _                   _  47c 

Of  the  five  Regular  Bodies        -                  -  477 

47.  To  measure  a  Tetracdron           -                   _                  _  47C 

48.  To  measure  an  Octaedron           _                  _                   „  47C 

49.  To  measure  a  Dodecaedron                            -                   -  478 
■tO.  To  measure  an  Eicosiedron                            -                   -  478 

51.  To  gauge  a  Cask                          _                   _                   _.  479 

52.  To  gauge  a  Mash  Tub                -                   -                   -  480 

53.  Having  the  Difference  of  Diameters,  Height  and  Content  of  a 

Mash  Tub,  to  find  the  Diameteis  at  Top  and  Bottom  480 

54.  To  ullage  a  Cask,  lying  on  one  side,  by  the  Gauging  Rod  481 

55.  To  find  a  Ship's  Tonnage  _  _  ^  481 
The  Proportions  and  Tonnage  of  Noah's  Ark  -  -  481' 
Questions  in  Mensuration  _  _,  _  48.. 
Book  Keeping  by  Single  Entry  _  _  -  488 
%- ~ — ^    Double  Entrv                -                   -^                   «->  505 


EXPLANATION 

OF   THii    CHARACTERS   tiADE    USE   01"   IN   THIS   TREATISB. 

:--  The  sign  of  equality  :  as  12  pence  =  1  shilling,  signifies  that  1.2  pence  n.T0 
ft'qual  to  one  shilling  ;  and,  in  general,  that  whatever  precedes  it  is  equal  to  what 
follows. 

+  The  sign  of  Addition :  as  5-f-5— 10,  that  is,  5  added  to  5  is  equal  to  10. 
Read  5  plus  5,  or  5  more  5  equal  to  10. 

—  Tl)^.  sign  of  Subtraction  :  as,  12 — 4=8,  that  is,  12  lessened  by  4  is  equal 
to  8,  or  4  from  12  and  8  remains.     Read  12  minus  4,  or  12  less  4  equal  to  8. 

X  The  sign  of  Multiplication  :  as  6  X  5=30,  that  is,  6  multiplied  by  5  is  equal 
to  30.     Read  6  into  5  equal  to  30. 

-Tt  or  5)30(  The  sign  of  Division  :  as  30-f-5=6,  that  is,  30  divided  by  5  is 
equal  to  6.     Read  30  by  5  equal  to  6. 

17 "" 

— ~  Numbers  placed  fractionwise,  do  likewise  denote  division,  the  numerator 

or  upper  number  being  the  dividend,  and  the  denominator  or  lower  number,  the 

divisor ;    thus,  —  is  the  same  as  875-^25=35. 
2b 
:  ::  :  The  sign  of  proportion,  thus,  2  :  4  ::  8  :  16,  that  is,  as  2  is  to  4  so  is  S  to  1^. 
~  Signifies  Geometrical  Progression. 

9 — 24-6=il3  Shews  that  the  difference  between  2  and  9  added  to  6  is  equat 
to  13.  Read  9  minus  2  plus  6  equal  to  13.  And  that  the  line  above  (called  a 
Ftnculimi)  connects  all  the  numbers  over  which  it  is  drawn. 

12 — 34-4=:5  Signifies  that  the  sum  of  3  and  4  taken  from  12  leaves  or  is  equal 
tV  5. 

I "    Signifies  the  second  power,  or  Square, 

I  ^    Signifies  the  third  power,  or  Cube. 

— !"^      .      .  — 'o 

I       Signifies  any  power  in  general,  as  6l2=rsquare  of  6  ;  and  50{   =:cUDe 

*jf  50,  (fee.  thus  m  signifies  either  the  square  or  cube,  or  any  other  power. 

v'l  or  i^  Prefixed  to  any  number  or  quantity,  signifies  that  the  square  root 
oi'  that  number  is  required.  It  likewise  (as  also  the  character  for  any  other 
root)  stands  for  the  expression  of  the  root -of  that  number  or  quantity  to  which 


it  is  prefixed.     As  ^36=6,  and  ^108-f  36=12,  and  36 1 2=6,  &c. 

3 i_ 

y/f  or  Is  Prefixed  to  any  number,  s'gnifies  that  the  cube  root  of  that  number 

is  required,  or  expressed.     As  y/216=6,  and  >/oiaH-216=9,   &c.  or  216 j^  = 
6,  &c. 

IV  "       I  n  — i  — i 

/    t)i'  l_-    Signifies  any  root  in  general.     As  301 2=2:squarQ  root,2l6i  s^^eube 
v        |m       ^  °  '  * 

n    ,     , 
root,  (fcc.      Thus,  —  signifies  either  the  square  root,  cube  root,  or  any  oth 

root  wh;#ever. 

ah  cd  When  streral  letters  are  set  together,  they  are  supposed  to  be  multi- 
plied into  each  other;  as  those  in  the  m,ai-gin  are  the  same  as  aXbXcXd,  and 
pc]:>resent  the  continual  product  of  quantities  or  nttmbers. 

1  ft  .  .  b 

—  Is  the  ro'ipr'ocal  of  rr,  and  —  is  the  reciprocal  of  — . 
//  h  a 

If  a  be  (he  root,  tlsrn  aXcc=aa  or  «2  is  the  square  of  a,  and  aXaXct=aaa 
or  (7  3  is  the  cube  of  r ,  iJcc. 

.Yote.     The  figure  above  is  called  the  index  of  the  power. 

It  is  usual  to  write  shillings  at  the  left  hand  o^  a  stroke,  and  pt nee  at  the 
right ;  thus,  13/4  is  thirteen  shillings  and  four  pence. 

.!\'ofe.  The  use  of  these  charactci-s  must  be  perfectly  inders1:(»d  ly  tlie  pu- 
T'jI.  as  he  may  have  occasion  foi:  \hcm'.. 


NEW  AND  COMPLETE 

SYSTEM  OF  ARITHMETICK. 


AHITIIMETICK  is  tlie  Art  or  Science  of  computing  by  num- 
bers, and  consists  both  in  I'heory  and  Practice.  The  Theo- 
ry considers  the  nature  and  quality  of  number?,  and  demonstrates 
the  rea'ion  of  praciical  operations.  The  l^ractice  is  that,  which 
•^hews  the  method  of  working:  by  numbers,  so  as  to  be  most  useful 
and  'expeditious  lor  business,  and  is  comprised  under  five  principal 
or  fundamental  Kules,  viz.  Notation  or  Numeration,  Addition, 
Subtraction,  Multiplication,  and  Division  ;  the  knowledge  of 
which  is  so  necessary,  that  scarcely  any  thing  in  life,  and  nothing 
in  trade  can  be  done  without  it. 


NUMERATION 

1.  TEACH CS  the  different  value  of  figures  by  their  different 
places,  and  to  read  or  write  any  sum  or  number  by  these  tenchar^ 
acters,  0,  1,  2,  3,  4,  5,  b\  7,  0,  9. — 0  is  called  a  cypher,  and  all  the 
rest  are  called  figures  or  digits.*  The  names  and  significations  of 
these  characters,  and  the  origin  or  generation  of  the  numbers  they 
stand  for,  are  as  follow  ;  0  nothing  ;  1  one,  or  a  single  thing  called 
an  unit;  1 +  1=2,  two  ;  2-}- 1=S,  three  ;  3-1-1=4,  lour  ;  4+1=^5, 
iive  ;  o-|-l=6,  six;  6-F  1=7,  seven  ;  7-}- 1=8,  eight ;  84-1=9, 
nine;  9-|-l  =  10  ten  ;  which  has  no  single  character  ;  and  thus,  by 
the  continual  addition  of  one,  all  numbers  are  generated. 

2.  The  value  of  figures  when  alone,  is  called  their  sm/)/g  value, 
and  is  invariable.  Besides  the  simple  value,  they  have  a  local  val- 
ue, that  is,  a  value  which  varies  according  to  the  place  they  stand 

*  VVitie  figures  or  digits  were  ol>taIned  from  the  Arabians,  and  were  introduc- 
ed into  Europe  in  the  ninth  century.  The  Arabs  probably  derived  the  deci- 
j»al  notation  from  hiJia.  Tlie  fcxagefimal  divifion  had  previoufly  been  in  gen- 
<:ral  ufe  in  Europe.  This  mode  of  divifion  is  yet  retained  in  a  few  cafes,  as  in 
the  divifion  of  time,  whcrt; ftxty  minutes  make  an  hoMT,ftxty  feconds  a  minute, 
&c.  The  figures  are  doubtlefs  called  digits  from  digitus,  a  fin^^er,  hecaufc  cotmf- 
ing  ufcd  to  he  performed  on  the  fingers. 

V 


18  iNUMEllATION. 

in  when  counecteil  together.  In  a  combination  of  figures,  reckon^ 
ing  from  the  right  to  the  left,  the  figirre  in  the  first  place  represents 
its  simple  value  ;  that  in  the  second  place,  ten  times  its  simple  val~ 
ue,  and  so  on  ;  each  succeeding  figure  being  ten  limes  the  value 
of  it  in  the  place  immediately  preceding.  There  is  no  reason  in 
the  nature  of  numbers  that  their  locaF  value  should  vary  according 
to  this  law.  They  might  have  been  made  to  increase  in  3,  4,  5, 
&c.  fold,  or  in  any  other  ratio.  The  tenfold  increa'se  is  assumed 
because  it  is  most  convenient. 

3.  The  values  of  the  places  are  estimated  according  to  their  or- 
der :  The  first  is  denominated  the  place  of  units  ;  the  second,  tens  ; 
the  third,  hundreds  ;  ^an(l  yo,opj  as  in  the  table.  'J'hus  in  the  num- 
ber— 5293467  ;  7,  in  the  first  place  signifies  only  seven  ;  6,  in  the 
second' placeV  s'!<>'Difi€g  6' tens;  of  sixty  ;  4,  in  th«  third  place,  four 
hundred;  3,  in' the'  fourth  place,  three  thousand;  9,  in  the  fifth 
place,  ninety  thoiisand  ;  2,  in  the  sixth  piace,  two  hundred  thou- 
sand ;  5,  in  the  seventh  place,  is  five  millions  ;  and  the  whole,  tak- 
en  together,  is  read  thus  ;  five  millions,  two  hundred  and  ninetv 
three  thousand,  four  hundred  and  sixty  seven. 

The  process  of  Numeration  may  be  more  clearly  seen  by  th^ 
following 

TABLE, 


tn 


tfi 


o  .2 

f.i  <!.! 

jn  15  w>    £      .                       '-A 

^  ,          •  'TO           cC                        '^3 

-g   w.  ^  3  S2  ^  =  .2             -  if' 

fn   *-  'O    t»  .J3  ti?  ~  -t3    m  c          >«  ~  -o    'i^ 

•X3     en     ~  "w     cfl  O  "3     tf)     S  "w  t«  .2  "^     fi     "  "^     (A    "'' 

M2r-~XE^WEHE-"2:r-^S:  —  H^I:•^:_• 
2  4  5,  9  3  8.  6   7   5,  2  6   7.  {5   9    1,3  4  d: 


Six  places  of  figures,  beginning  on  the  right,  are  called  a  period, 
and  each  successive  six  places  another  period.  Each  period  is  con- 
sidered as  divided  into  two  half  periods  of  three  figures  each.  These 
are  distinguished  by  the  comma,  antl  the  point  for  a  period.  There 
is  an  obvious  reason  for  this  division  into  periods,  for  at  the  begin- 
ning of  each  period,  there  is  anew  denomination  of  units,  of  which 
the  tens,  hundreds,  thousands,  fiic.  are  numerated  as  in  the  fir^ 
period. 

4.  A  cypher,  though  it  is  of  no  signification  itself,  yet,  it  pos- 
sesses a  place,  and,  when  set  on  the  right  hand  of  figures,  in  whole 


NUMERATION.  19 

numbers,  increases  their  value  in  the  same  tenfold  proportion  ;  thus, 
9  signifies  only  nine  ;  but  if  a  cypher  is  placed  on  its  right  hand, 
thus,  90,  it  then  becomes  ninety  ;  and,  if  two  cyphers  be  placed  on 
its  right,  thus,  900,  it  is  nine  hundred,  &c. 

6.  To  icnuijierate  any  parcel  of  figures,  observe  the  following 
Knle. 

First,  :Committhe  words  at  the  head  of  the  table,  viz.  unit?,  tens, 
hundreds,  Lc  U)  memory,  then,  to  Ihe  simple  value  oi  each  figure, 
join  the  name  of  its  place,  beginning  at  the  I'^it  hand,  and  reading 
towards  the  right.— More  particularly — I.  Place  a  dot  under  the 
right  hand  tigure  of  the  2tl,  4th,  Gth,  8th,  &c.  half  period?,  and  ihe 
figure  over  such  dot  will,  universally,  have  the  name  of  thousands. 
— 2.  Place  the  fiijures,  1,2,  3,  4,  &c.  as  indices  over  the  2d,  3d. 
i'th,  &,c.  period.  These  indices  will  then  shew  the  number  of  times 
the  millions  are  incrt^ased. —  Ihe  figure  under  1,  bearing  the  name 
of  millions,  that  under  2,  the  naipe  of  billions  (or  millions  of  mill- 
ions) that  under  3,  trillions. 

EXAMPLE. 

Scxtlllions.  Quintilli.   Quatrill.    Trillions.    Billions.    Millions.      Units. 
rvA-*n      Ow^--^      ro*-^      r^A-^      n^/^o      rvA-o      r^^.^-n 
th.  un.       th.  un.       tli.  un.       th.  un.       th.  un.      tU.  uu.    c.x.t.c.x.\j. 
r*/-nfV«*^  r».««of^^^>  r^^^^rx-^^^  r»-»«<» f^^-<i  f»»A^^r^*-n  (>/^^r>^'^\  t'K\^fs^.^^ 

6  5  4  3  2  1 

:}13,'208,000,341, 620,057,219,356,809,379,120,406,1 29,763 

H  H  H  H  H  H  H 

s-  =r  D-  sr  cr  tr  zr 

o  o  o  o  o  o  o 

1=  c  c  c  a  a  c 

c/i  :n  ro  »  en  (u  rn 

w  V  V  to  at  »  (»  « 

3  3  D  D  a  o  a 

C«  CLi  Crf  Ci  !^  CU  CU 

CA  en  CA  (fj  ci<  tn  en 

NoTis  1.  Billions  is  substituted  for  millions  of  millions  :  Trillions, 
for  millions  of  millions  of  million*  ;  Q,uatrillions,  for  millions  of  mill- 
ions of  millions  of  millions,  and  so  on. 

These  names  of  periods  of  figures,  derived  from  the  Latin  nume- 
ral*!, may  he  continued  without  end.  They  are  as  follows,  for 
twenty  periods,  viz.  Units,  Millions,  Uillions,  Trillions,  Quatrill- 
ion«,  Quiutillions,  Sextillions,  Septillions,  Octitlions,  Nonillions,  Do- 
minions, L^idecillions,  Duodecillions,  1  redecillions,  Q,uatuordecili  • 
ions,  Quindecillions,  Sexdecillions,  Septendecjllions,  Octodecillion-', 
Noremdecillions,  Vigintil lions. 

The  Application. 

Wrhe  (iQwn,  in  proper  Jigures,  ihefoUazmitig  numbers. 

Fifteen.  -- ^i 

Two  hundred  and  feyenty  nine.  -------       279 

Three  thoufand  four  hundred  and  three.     -  -         -         -         -         - 

'i'hirty  fevcnthoufand,  five  hundred  and  fixty  fcvcn,    -         -         -  375^7 

Four  hundred,  one  thoufand  and  twenty  eight.  _         -         - 

Nine  millions,  ftvcnty  two  ti)Ouf^nd  snd  two  hundred.         -         -        907x200 
rifty  five  millions,  three  hundred,  nine  thoufand  and  nine.    - 
Eight  hundred  millions,  forty  four  thoufand,  and  fifty  five. 
Two  thoufand, five  hundred  and  forty  three  millions,  four?  a?414^l"OJ 

hundred  and  thirty  one  thoufand,  feven  hundred  and  two.  5  .    <    J   /    * 


20  SIMPLE  ADDITION. 

Write  down  in  words  at  length  the  following  numbers. 

8  437  709040  3476194  7584397647 

17  3010  879066  84094007  49163189186 

129  76506  4091875  690748591  500098400700 

Notation  by  Roman  Letters. 

I.  One.  XV.  Fifteen.  CC.  Two  hundrea. 

H.   Two.  XVI.  Sixteen.         CCC.  'Jl^ree  hundre^L 

Ilf.   Three.  XVII.   Seventeen.  CCCC.   Four  hundred. 

IV.  Four.  XVIll.  Eighteen.  D.  or  Ij-  Five  hundred. 

V.  Five.  XIX.   Nineteen.      DC.  Six  hundred. 

VI.  Six.  XX.  Twenty.  l)CC.  Seven  hundred. 

VII.  Seven.  XXX.  Thirty.         DCCC.  Eight  hundred. 

VIII.  Eight.  XL.  Forty.  DCCCC  Nine  hundred. 

IX.  Nine.  L.  Fifty.  M.  or  €13.  One  Thousand. 

X.  Fen.  LX.  Sixty.  ^DO-   I'^ive    Fhousand. 

XI.  Eleven.  LXX.   Seventy.       ^OOO-  Fifty  thousand. 

XII.  Twelve.       LXXX.   Eighty.      I OOO'O  )3     Five  hund.  thou 

XIII.  Thirteen.   XC.   Ninety.  MDCCCV'llI.  One  Thousand, 

XIV.  Fourteen.    C.  Hundred.  eight  hundred  and  eight. 

A  less  literal  number  placed  after  a  greater, always  augments  the 
value  of  the  greater  ;  if  put  before,  it  diminishes  it.  1  hus,  VI.  is 
6  ;  IV.  is  4  ;  XI.  is  11  ;   IX.  is  9,  &c. 

The  practice  of  counting  on  the  lingers  doubtless  originated  the 
method  of  Notation  by  Roman  Letters.  The  letter  I  was  taken 
ibr  one  finger,  or  one  ;  and  hence  11,  for  two  ;  111,  tor  three  ;  Illl, 
ibr  four;  and  V,  as  representing  the  opening  between  the  thumb 
and  fore  finger,  and  bfing  also  an  easier  combination  of  the  marks 
for  the  fingers,  was  taken  for  five.  As  IV  is  a  sim[)ler  expression 
for  four  than  the  above,  it  was  doubtless  adopted  lor  this  reason, 
and  on  the  general  principle  too  that  a  less  literal  number  placeil 
before  a  greater  shotild  diminish  the  greater  so  much,  and,  plaroj 
after  a  greatershould  augment  it  so  much.  Hence  as  IV ,\>fovr  ; 
VI  is  six  ;  Vlll  is  eighty  and  so.  on.  Ten  was  expressed  by  X,  lie- 
cause  it  \s  two  V^s  united,  and  tvyice  five  is  ten.  i'^ifu  was  express- 
ed by  L,  because  it  is  half  of  C  or  E,  as  it  was  anciently  written, 
and  C  is  the  initial  of  the  Latin  centum.,  one  hundred. 

Five  hundred  is  expressed  by  D,  because  it  is  half  of  the  Goth; 
>C  CD  or  M.,  the  initial  oi^  m.ilU,  one  thousand. 


ADDITION 

iS  il;e  puttina:  together  of  two   or  more  numbers,  or  sum?,  to 
make  them  one  total,  or  whole  sum. 

SIMPLE  ADDITKIY 
Is  the  adding  of  several  integers  or  whole  numbers  together, 
which  are  all  of  one  kind,  or  sort  ;  as  7  pounds,  12  pounds,  and  '10 
poun(j^;.Jieing  added  '.oget.herj  their  aggregate,  or  sum  total,  i^  ."V 
pounds.*  ; 


SIMPLE  ADDITION.  2i 

Rule. 
llavinf^  placed  units  under  units,  tens  under  lens,  &;c.  draw  aline 
underneath,  and  begin  with  the  units;  after  adding  up  every  iig- 
ure  in  that  column,  consider  how  many  tens  are  contained  in  their 
sum,  and  placing  the  excess  under  the  units,  carry  so  many  as  you 
have  tens,  to  the  next  column,  of  tens  :  Proceed  in  the  same  man- 
ner through  every  column,  or  row,  and  set  down  the  whole  amount 
of  the  last  row.* 

*  This  Rule  as  well  as  the  method  of  proof,  is  founded  on  the  known  axiom, 
"  the  whole  is  equal  to  the  fum  of  all  its  parts."  The  method  of  placing  the 
numbers,  and  carrying  for  the  tens,  is  evident  from  the  nature  of  notation  ;  for, 
any  other  difpofition  of  the  numbers  would  alter  their  value ;  and  carrying  one, 
for  every  ten,  from  an  inferiour  to  a  fuperiour  column,  is,  evidently,  right;  be- 
canfc  one  unit  in  the  latter  cafe  is  equal  to  the  value  often  units  in  the  former. 

Kefides  the  method  of  proof  here  given,  there  is  another,  by  carting  out  the 
nines  ;  thus  : 

X.  Add  the  figures  in  the  upper  row  together,  and  find  how  many  nines  are 
contained  in  their  fum. 

2.  Reject  the  nines,  and  fet,  down  the  remainder,  directly  even  with  the  fig- 
ures in  the  row. 

3.  Do  the  fame  with  each  of  the  given  numbers,  and  fct  all  the  exccfles  of 
nines  in  a  column,  and  find  their  fum  ;  then,  if  the  excefs  of  nines  in  this  fum, 
found,  as  before,  is  equal  to  the  excefs  of  nines  in  the  fum  total ;  the  queftion 
is  fuppofed  to  be  right. 

Example. 

573      u    5"J  This  method  depends  upon  a  property  of  the  number 

^  I  9,  which, except  3,  belongs  to  no  other  digit  whatever; 

r  viz,  that  any  number  divided  by  9,  will  leave  the  fame 

^  1  remainder,  as  the  fum  of  its  figures, or  digits,  divided  by 


9456 
8471 
53*4 


^^6891  »<     6j     9:  which  may  be  thus  demonftrated. 

Demonflration.   Let  there  be  any  number,  as  5432;  this,fcparated  into  itsfeveral 

parts,  becomes  5000  f4COf  30  i  a;  but  5000 -5 Xi 000 --.5X999-1-1  —  5X999 

-j-5.     In  like  manner  400     4     99  f  4,  and  30— 3x9  '  3.     Therefore,  543*  — 

5x999     5»    :  4     99-f4»+3     9  *_3j:*=J- 999-1-4X99  1-3  X9r5-f4H-3H- 3- 

543*     5x999  1-4x99  f  3  X9-!-5H-4+3-1  a 

And u^ ■ ;  but  5  X999-f4X99-h3  X9  is 

.9  9 

divilible  by  9  ;  therefore, 5432,  divided  by  9,  will  leave  the  fame  remainder,  as 
5-i  4  '-34-*>  divided  by  9;  and  the  fame  will  hold  good  of  any  other  number 
whatever. 

The  fame  property  belongs  to  the  number  3  :  However,  this  inconveniency 
attends  this  method,  that,  although  the  work  will  always  prove  right,  when  it  is 
l"o  ;  it  will  not,  always,  be  right,  when  it  proves  fo  ;  I  have, therefore,  given  this 
demonflration  more  for  the  fake  of  the  curious,  than  for  any  real  advantage. 

In  cafling  out  the  nines,  proceed  thus.  Begin  with  the  uppermoft  row  of  the 
r,xample  at  the  left  hand  ;  5  and  7  are  la,  from  which  take  out  nine,  and  3  re- 
mains :  3  added  to  3  make  6,  which  muft  be  added  to  the  8,  becaufe  6  is  lefs 
than  9, and  the  fum  is  14  :  caft  out  nir\e  and  5  remains,  which  is  to  bt  placed 
at  the  right  againfl  the  row,  as  in  the  example.  In  the  next  row,  9  the  firft  fig- 
ure, may  be  omitted  becaufe  it  is  9  ;  then  1  and  5  make  6,  which  added  to  the  6, 
make  1  a,  from  which  take  out  9.  and  3  remains  to  be  placed  on  the  right  of  the 
row  as  liefore.  Proceed  thus  with  all  the  rows  and  with  the  fum  at  bottom. 
Then  add  the  remainders  ajainfV  thefevrral  ro^ys,  cafting  out  9  as  often  as  i: 


22  SIMPLE  ADDITION. 

Proof.  Begin  at  the  top  of  the  sum  and  reckon  the  figures  down; 
wards,  in  the  same  manner  as  ihey  were  added  upwards,  and,  if  it 
be  right,  this  aggregate  will  be  equal  to  the  first.  Or,  cut  off  the 
upper  line  of  figures,  and  find  the  amount  of  the  rest;  then,  if  the 
amount  and  upper  line,  when  added,  be  equal  to  the  sum  total,  the 
work  is  supposed  to  be  right. 


Addition  and  Subtraction  Table. 


j  1  1  2 

1   - 

4 

1   ^ 

R  ' 

7 

8 

1   • 

10 

12 

|2 
3 
~4 
~~b 
~6 

'"9 

To 

4 

i   ^ 

^ 

1   '7 

8 

9 

10 

1  >> 

1  '^  i 

13 
14 

~i5 

5 

1   ^ 

' 

1   « 

9 

10 

11 

1  1 2 

1  .1-3 

8 

1   '^ 

8 

1   9 

10  1 

" 

12 

13 

14  ' 

1  -. 

16  1 

7 

1   ^ 

9 

10 

" 

12 

13 

1  1  <!- 

•5  1 

li'   ; 

i "  1 

8 

1   9 

10 

1 1 

1" 

13  1 

1* 

1  1''^ 

16    1 

17 

I8| 

9 

1  10 

n 

1  5-' 

13 

14 

lu 

1- 

17  1 

IH  1 

19  1 

10 

I  11 

12 

i  13 

H 

'• 

16 

1  1  '^ 

18  1 

!-  ! 

■■<   1 

II 

12 

1  12 
1  13 

13 

1  l^l 

16  1 

le  1 

17 

iH  1 

19  1 

'■Z>-'    1 

■;i  i 

14 

15  1 

16  1 

17  1 

18 

11:* 

20  1 

":   !  1 

2 

When  you  would  add  two  numbers,  look  one  of  them  in  the  left 
hand  column  and  the  other  at  top,  and  in  the  common  angle  of 
meeting,  or,  at  the  right  hand  of  the  first,  apd  under  the  second, 
you  will  find  the  sum — as,  6  and  8  is  13. 

When  you  would  subtract  :  Find  the  number  to  be  subtracted  in 
t!ie  left  hnn<l  column,  run  your  eye  along  to  the  right  hand  till  you 
im(\  the  number  from  which  it  is  taken,  and  right  over  it  at  top  you 
Will  find  the  difference — as  8,  taken  from  13,  leaves  5. 


trcrurs,and,  if  the  remainder  be  the  fame  as  that  againfl:  the  fum.as  it  is  in  this 
example,  tlie  work  is  prefumed  to  he  right. 

An  eaTier  method  of  catling  out  the  nines,  is  to  begin  as  before,  and  when  the 
fum  exceeds  nine,  to  add  the  fioiircs  themfcives  of  this  lum  as  before,  and  fo  pro- 
tccd,  and  this  new  fum  will  always  l)e  equal  to  the  remainder  after  nine  is  ta- 
ken from  the  firft  fum.  Thus,  as  before,  3  and  ^  are  i2, — now  add  the  num- 
bers of  this  fum,  which,  being  i  and  2,  make  3,  equal  to  the  remainder  after  9 
js  taken  from  12;  then  3  and  3  added  to  8  make  14,— vidd  the  i  and  4, and  the 
fum  is  5,  the  fame  as  the  remainder  above.  In  the  next  row, — omitting  the  9, 
t!ie  fum  is  n,  the  numbers  of  which,  i  and  a,  make  3,  the  remainder  as  above, 
rhe  fame  will  hold  true  in  any  cafe. 

Note.  It  fliould  be  noticed  that  the  method  of  proof  for  this  rule,  and  vari- 
ous others,  depends  upon  the  accuracy  of  both  operations.  It  does  not  follow 
heaufc  the  refult  is  the  fame  by  both  operations,  that  there  can  be  no  error.  For 
?joth  operations  may  be  incorrectly  performed,  and  the  refults,  though  alike, 
t  rroncous.  The  bejl  proof  that  any  rtfult  is  rigfit,  is  \.\\z  correct ^erf'.rmance  of  all 
c/;e  cferai}cr,\ 


SIMPLE  ADDITION 


23 


E^ 

iAMPLES. 

1. 

2, 

3. 

4. 

5. 

6. 

/:• 

ft. 

Cvvt. 

Miles. 

Yards. 

£. 

1 

12 

123 

1234 

12345 

987654321 

2 

34 

456 

5678 

67890 

123456789 

3 

56 

785 

9098 

98765 

234567891 

4 

78 

12 

7654 

43210 

345678910 

6 

90 

345 

3210 

12345 

456789123 

6 

1 

678 

€2 

67890 

567879287 

7 

23 

901 

4713 

74100 

678900028 

8 

45 

234 

131 

64786 

789400690 

9 

67 

567 

9128 

19876 

548769138 

3nm,  45 

40908 

In  the  first  Example,  the  student  adds  together  the  several  num- 
bers, and  finds  the  sum  to  he  45  ;  and,  as  there  is  but  one  column, 
he  must  set  down  45  for  the  answer. 

In  the  4lh  Ex.  the  student  will  add  the  numbers  of  the  column  on 
the  right  hand,  which  he  will  find  to  be  38  ;  he  will  set  the  8  un- 
der the  column,  and  carry  3  to  the  next  column.  The  next  col- 
umn with  the  3  to  be  carried,  he  will  find  to  be  40  ;  he  must  set 
down  the  0,  and  carry  4  to  the  next  column.  This  will  be  found 
to  be  29 ;  the  9  is  to  be  set  under  the  column,  and  the  2  carried 
to  the  next  column,  which  makes  40;  the  cypher  is  to  be  put  un- 
der the  column,  and  the  4  will  take  the  next  higher  place,  for  it 
is  evident  the  whole  must  be  set  down.  The  same  course  must 
be  pursued  in  each  example. 


7. 

8. 

9. 

10. 

1234567 

1234567 

67 

1234567 

2345678 

723456 

123 

9876543 

3456789 

34565  • 

4567 

2102863 

4567890 

4566 

89093 

4321234 

5678209 

333 

654321 

5682098 

6789098 

90 

1234567 

6543218 

1997577 

11.  What  is  the  sum  of  3406,  7980,  345  and  27  ?     Ans.   1175tJ. 

12.  A  man  borrowed  of  his  neighbour,  thirty  dollars  at  one  time, 
one  hundred  and  66  at  another,  and  seventy  Cive  at  another:  how 
much  did  he  borrow  in  the  whole  ?     Ans.  271  dolls. 

13.  Four  boys  collected  ehesnuts ;  A.  had  4096,  B.  16784,  C. 
11590,  and  D.  fi?^  hundred  and  57  ;  how  many  were  there  in  the 
whole  ?     Ans.       . 

14.  Four  boys,  on  counting  their  apples,  found  that  A.  had  67, 
B.  11  more  than  A,  C.  had  101,  and  D,  had  sixteen  more  than  C,. 
how  many  had  they  all? 

15.  rhe  Deluge  happened  2348  years  before  the  birth  of  our 
Saviour,  and  America  was  discovered  1492  years  alter  it;  bow- 
many  years  intervened  ^    ': 


IMPLE  SUBTRACTION 


SUBTRACTION. 

TEACHES  to  take  a  less  number  from  a  greater,  to  fjnti  aihlrcj 
sliewinf!^  the  inequality  or  flifferet)ce  between  the  given  numbers. 
'J'he    (greater  number  is  called  the  Minuend.      The  less  number  in 
railed  (he  Subtrahend.     The  ilitlerence,  or,  what  is  left  after  the 
subtraction  is  made,  is  called  the  Remainder. 

SIMPLE  SUBTrIcTIPN 

Teaches  to  find  the  diflerence  between  any  two  numbers,  which 
are  of  the  same  kind. 

Kile. 
Place  the  larger  number  uppermost,  and  the  less  undernealh,  so 
that  units  may  stacid  under  units,  tens  under  tens,  &.c.  then,  draw- 
ing a  line  underneath,  begin  with  the  units,  and  subtract  the  lower 
from  the  upper  figure,  and  set  down  the  remainder  ;  but  if  the  low- 
er figure  be  greater  than  the  upper,  add  ten,  and  subtract  the  low- 
er figure  therefrom  :  To  this,  diifer-e nee,  add  the  upper  figure, 
Avhich  being  set  down,  you  must  add  one  to  the  ten's  place  of  tiie 
lower  line,  for  that  which  you  added  before  ;  and  thus  proceed 
ihrOugh'the  whole.'* 

Proof. 
In  either  simple,  oi*  compound  Subtraction,  add  the  remainder 
and  the  less  line   together,  whose   sum,  if  the  work  be  right,  will 
be  equal  to  the  greater  line  :  Or  subtract  the  remainder  from  the 
greater  litse,  and  the  difference  will  be  equal  to  the  less. 

Examples. 

3.  4. 

Miles.         Yards. 

4670         58934 

4020  6182 


1. 

o. 

i;. 

i-'- 

From  125 

305 

Take  12 

103 

Kern.     13 

Proof.  i^5 

6. 

6. 

Feet. 

Cwt. 

879047 

9187641 

164348 

91843 

58934 


*  jDc'w.  When  all  the  figures  of  the  lefs  number  arc  Icfs  tirin  their  corres- 
pondent figures  in  the  greater,  the  difference  of  tlie  fij)ui  ts,  in  the  feveral  like 
places,  muft,  all  taken  together,  make  the  true  difference  fought  ;  bccuufe,  is 
the  funi  of  the  parts  is  equal  to  the  whole  ;  fo  muft  the  ium  of  the  differences, 
i>f  all  the  fimilar  parts,  he  equal  to  the  difference  of  the  whole. 

1,  When  any  figure  in  the  greater  number  is  lefs  than  its  correfpondcnt  fi;;- 
lire  in  the  lefs,  the  ten  which  is  added  by  the  Rule,  is  the  value  of  an  unit  in 
t'le  next  higher  plice,  l)y  the  nature  of  notation;  and  the  one  which  is  adilcd 
to  the  next  place  of  the  lefs  number,  is  to  dinilnifli  the  correfpondcnt  place  oi 
». he  greater,  accordingly  ;  which  is  only  taking  from  oiic  place  and  adding  at^ 
irmch  to  another,  wlicrcby  the  total  is  never  changed  :  And,  by  this  means, th:: 
greater  is  refolved  into  fiich  parts,  a*  are,  eacli,  greater  than,  or  equal  to,  the 
1'.inil;-ir  pjrt  of  the  Itfs ;  ;ind  rhc  difference  of  the  correfpondcnt  figufC8,  taken 
together,  will,  evidently,  make  up  the  difference  of  the  whole. 

The  truth  of  the  rr.ethod  of  proof  is  evident ;  lor  tlie  difTcrercc  of  two  nuir- 
Iv-r^  -.iddcd  ti.  tlu-  I(..'V,  !«  nntiifoft.'v.  equ.il  to  the  greater. 


SIMPLE  MULTIPLICATION.  25 

The  operation  on  the  first  three  Examples  is  Fufficiently  plain. 
In  the  4th  Ex.  I  begin  on  the  right  hand,  and  lake  2  from  4,  and 
set  down  the  difference  2,  under  the  column.  As  8  is  greater  than 
.3,  I  add  10  to  3,  which  makes  13,  and  from  it  take  the  8,  and  6  is 
the  diflerence  to  be  set  down.  As  I  add  10  to  the  3,  I  now  add  1 
to  the  1  in  the  next  higher  place,  because  10  in  one  place  is  equal 
only  to  1  in  the  next  higher  place,  and  take  the  2  from  the  9,  and 
the  difference  is  7.  The  rest  of  the  work  is  obvious.  The  same 
proofs  must  be  followed  in  every  similar  case. 

7.  8.  9. 

100200300400500600700800900  10000  1000000 

980760.54032011023043067089  9999  1 


10.  What  is  the  difference  of  40875  and  38968?     Ans.  1907. 

11.  What  number  must  be  added  to  6892,  so  that  the  sum  shall 
be  8265?     Ans.  1373. 

12.  America  was  discovered  in  1492  ;  how  many  years  have 
elapsed  since  ? 

13.  If  you  lend  your  friend  3646  dollars,  and    afterwards  are 
paid  2998  dollars  ;  how  much  is  still  due  ?     Ans.  648  dollars. 

14.  If  a  man  was  seventy  live  years  old  in  the  year  1821,  in  what 
year  was  he  born  ?     Ans.  1746. 

15.  The  Independence  of  the  United  States  was  declared  July 
4th,  1776  ;  how  many  years  have  passed  since  ?     Ans. 

16.  Sir  Isaac   Newton  died  in  the  year  1727,  aged  eighty  five  ; 
in  what  year  was  he  born?     Ans.  1642. 


MULTIPLICATION 

TEACHES  to  find  the  amount  of  one  number  increased  as  many 
times  as  there  are  units  in  another,  and  thus  performs  the  work  of 
many  additions  in  the  most  compendious  manner  ;  brings  numbers 
of  great  denominations  into  small,  as  pounds  into  shillings,  pence 
or  farthings,  &c.  and,  by  knowing  the  value  of  one  thing,  we  find 
the  value  of  many. 

The  number  given  to  be  multiplied,  is  called  the  Multiplicand. 

The  number  given  to  multiply  by,  is  called  the  Multiplier. 

The  multiplicand  and  multiplier  are  otten  called /oc^or^. 

The  result  of  the  operation,  or  the  number  found  by  multiplying, 
is  called  the  Product. 

Multiplication  is  distinguished  into  Simple  and  Compound. 

SIMPLE  MULTIPLICATlok 

Is  the  multiplying  of  any  two  numbers  together,  withput  having 
regard  to  their  signification  ;  as  7  tiines  8  i^  56,  &r. 

D 


26  iSlMPLE  ML'LTIPLICATION. 

Multiplication  and  Division  Table. 


1 

2 

3 

4 

5 

6 

7| 

M 

9| 

10  1 

111 

12 

2 

4 

6| 

8| 

10  1 

12  1 

14  J 

16  1 

18  I 

20.1 

22  1 

24 

3 

6 

9 

12 

15 

18 

21  1 

24  1 

27  1 

30  1 

33  1 

36 

4 

8 

12 

16  1 

20 

24 

28  1 

32  1 

36  1 

40  1 

44  1 

48 

5 

10 

15 

20 

25 

30 

35 

40 

45  1 

50  1 

55  1 

60 

6 

12 

18  1 

24  1 

30 

36 

42  1 

48  1 

54  1 

60  1 

66  1 

72 

7| 

14  1 

21  1 

28  i 

35  1 

42  i 

49  1 

56  1 

63  1 

70  1 

77  1 

84 

8| 

16  1 

24  1 

32  1 

40  1 

48  1 

56  1 

64  1 

72  1 

80  1 

88  1 

96 

9 

18 

27 

36 

45  1 

54  1 

63  1 

72  1 

81  1 

90  1 

99  1 

108 

10 

20 

30  i 

40  1 

50  j 

60 

70  1 

80  1 

90  1 

100 

110  1 

120 

n  1 

22 

33 

44 

55  1 

66 

|77| 

88  ! 

99  1 

110  1 

121  1 

132 

12  1 

24 

36 

48 

60 

72 

84  1 

96  1 

108  1 

120  1 

132  1 

144 

To  learn  this  Table  for  Multiplication  :  Find  your  multiplier  hi 
the  left  hand  column,  and  your  multiplicand  at  toj),  and  in  the  com- 
mon angle  of  meeting,  or  against  your  multiplier,  along  at  the 
right  hand,  and  under  your  multiplicand,  you  will  find  the  product, 
or  answer. 

To  learn  it  for  Division  :  Find  the  divisor  in  the  left  hand  col- 
umn, and  run  your  eye  along  the  row  to  the  right  hand  until  you 
find  the  dividend  ;  then,  directly  over  the  dividend,  at  top,  you  will 
find  the  quotient,  shewing  how  often  the  divisor  is  contained  in  the 
dividend. 

Rule. 
Having  placed  the  multiplier  under  the  multiplicand  so  that  unit* 
stand  under  units,  tens   under  tens,  &c.  and  drawn  a  line   undei 
them,  then, 

1.  IVhcn  the  multiplier  does  not  exceed  12;  begin  at  the  right  hand 
of  the  multiplicand  and  multiply  each  figure  by  the  multiplier,  set- 
ting down  the  unit  figure  under  units,  and  so  on,  and  carrying  for 
the  tens  to  the  next  place,  as  in  addition,  and  the  work  is  done.* 

2.  When  the  multiplier  exceeds  12  ;  multiply  each  figure  of  the 
multiplicand  by  every  figure  in  the  multiplier  as  before,  placing 
the  first  figure  of  each  product  exactly  under  its  multiplier  :  then 

•  Dem.  When  the  multiplier  is  a  finglc  digit,  it  is  plain  that  we  find  the  pro- 
duiSl ;  for,  bymuhiplying  every  figure,  that  is,  every  part  of  the  multiplicand, 
we  multiply  the  whole  ;  and,  the  writing  down  of  the  products,  which  arc  lei>. 
than  ten,  ortheextefs  of  tens,  in  the  places  of  the  figures  multiplied,  and  car- 
rying the  number  of  tens  to  the  product  of  the  next  place,  is  only  gathering  tO' 
gether  the  limilar  parts  of  the  refpective  products,  and  is,  therefore,  the  fame, 
in  effect,  as  though  we  wrote  down  the  multiplicand  as  often  as  the  multiplier 
txprefies,  and  added  them  together;  for  the  fum  of  every  column  is  the  pro- 
duct  of  the  figures  in  the  place  of  that  column  ;  and  the  products,  collected  to- 
gether, arc  evidently  equal  to  the  whole  required  product. 


SIMPLE  MULTIPI^CATION. 


ad<^  together  these   several  |)roducts  as  they  stand,  an4  their  sum 
will  be  the  total  product.* 


*  If  the  multiplier  be  a  number,  made  up  of  more  than  one  figure;  after  we 
have  found  the  product  of  the  multiplicand  by  the  firft  figure  of  the  multiplier, 
38  above,  we  fuppofe  the  multiplier  divided  into  parts,  and,  after  the  fame 
manner,  find  the  product  of  the  multiplicand  by  the  fecond  figure  of  the  mul- 
tiplier; but  as  the  figure,  by  which  we  are  multiplying,  ftands  in  the  place  of 
tens,  the  product  muft  be  ten  times  its  fimple  value  ;  and;  therefore,  the  firft  fig- 
ure in  this  product  muft  be  noted  in  the  place  of  tens,  or,  which  is  the  fame, 
directly  under  the  figure  we  are  multiplying  by.  And,  proceeding  in  the  fame 
manner  with  all  the  figures  of  the  multiplier,  feparately,  it  is  evident  we  fliall 
multiply  all  the  parts  of  the  multiplicand  by  all  the  parts  of  the  multiplier; 
therefore,  thcfe  fcveral  products,  being  added  together,  will  be  equal  to  the 
whole  required  product. 

The  reafon  of  the  method  of  proof,  depends  upon  this  propofition,  that  if 
two  numbers  arc  to  be  multiplied  together,  cither  of  them  may  be  made  the 
multiplier  or  multiplicand,  and  the  product  will  be  the  fame. 

A  fmall  attention  to  the  nature  of  numbers  will  make  this  truth  evident ;  for 
5X9  =  45  =9xr,;  and,  in  general, 2X)X4X5X6,  &c.  ==  3X2X6  XoX-*,&c. 
without  any  regard  to  the  order  of  the  terms ;  and  this  is  true  of  any  number 
of  fadtors  whatever. 

The  following  examples  arc  fubjolncd,  to  make  the  reafon  of  the  rule  appear 
as  clearly  as  pofllblc. 

23795(> 
3728 

1903648  =  8  times  the  multiplicand. 

475912  r=  20  times  ditto. 
1665692  =  700  times  ditto. 
713868  =  3000  times  ditto. 


64753 
5 

15=         3X5 

25  =       50X5 

35     =     700X5 

20       =  4000X5 

30         =60000X5 

323  ;65=t]4753  X 5  887099068=3728  tiroes  ditto. 


Another  method  of  proving  the  rule  is  as  follows.     Let  the  factors  be  6475I; 
.md    ■>.     Now   ;)4753=OOU004-4000-1-7004-50-|-:.'.     The  fum  of  the  products 
of  thcfe  quantities  feverally  multiplied  by  5,  is  the  true  product.     Then 
60000-1- 4U00+700^-504-;3  is  one  faAOr.     5  the  multiplier  the  other  faaor. 
300000^-20000^-3500-f  250+  15=32'i765=64753  X  5. 
Or  let  the  faaors  be  45  and  24.     Then  45= 40-f  5,  and  24=20+4,  and 

40+5  multiplicand.  l.ct  the  favSlors  be  24  and  2^1.     Then, 

20+4  multiplier.  20+4 

800+100=45  X  20  '^^+-^ 

160+20=45X4  400+80  =24x20 

800+260+20=  1080=  15  X  2 1.     . .^^+16  =  24  X  4. 

400+160+16  =  576=24X21. 

Multiplication  may  alfo  be  proved,  by  cafting  out  the  nines  ;  but  is  liable  to 
the  inconvenience  before  mentioned. 

It  may  likewifc  be,  very  naturally,  proved  by  divifion  ;  for  the  produtSt,  be- 
ing divided  by  eitlier  of  the  fa(Slors,wIII,  evidently,  give  the  other  ;  and  it  might 
not  be  amifs  for  the  pupil  to  be  taught  divifion,at  the  fame  time  with  multipli- 
cation ;  as  it  not  only  fcrves  f'^r  proof;  but  alfo  gives  him  a  readier  knowledge 
of  them  both.  But  it  would  have  been  contrary  to  good  method  to  have  giv- 
en this  rule  in  the  text,  bccaufc  the  pupil  is  fuppofed,  as  yet,  to  be  unacquain'- 
ed  with  divifion. 


28  SIMPLE  MULTIPLICATION. 

Proof. 

Multiply  the  multiplier  by  the  multiplicand. 

Multiply  3851  by  3.  By  addition. 

3851  Multiplicand.  3851 

3  Multiplier.  3851 

3851 

11553  Product. 

11553  Sum. 

Having  placed  the  numbers  according  to  the  rule,— then  say,  3 
times  1  is  3,  and  place  3  directly  under  units  ;  then  3  times  5  is  15, 
set  down  5  and  carry  the  one  to  the  next  product.  Then,  3  times 
8  is  21,  to  which  the  1  is  to  b^  added,  making  25  ;  set  down  5  and 
carry  2.  Then  3  times  3  is  9,  and  the  2  to  be  carried,  make  11, 
which  set  down,  and  the  work  is  done.  The  result  is  thcsame  as 
is  obtained  by  addition. 

Multiply  6053  by  11. 
11 

Prod.  66583 

Proceeding  as  before,  multiply  3  by  II,  and  of  the  product,  33, 
set  down  3  under  units,  and  carry  3  ;  then  5.by  11,  and  to  the  pro- 
iluct,  55,  add  the  3  to  be  carried,  set  down  8,  and  carry  5 ;  then  O 
by  1,1,  and  as  the  product  is  0,  set  down  the  5,  which  was  to  be 
carried  ;  then  6  by  U,  and,  as  there  is  none  to  carry,  set  down  the 
product,  66,  and  the  operation  is  finished. 

Multiply  67013  by  29. 
67013  MuItipHcand. 
29  Multiplier. 


603117  Product  by  9,  the  units  of  the  multiplier. 
134026    Product  by  2,  the  tens  of  the  multiplier. 

1943377  Product  or  answer. 

in  this  example,  the  multiplicand  is  first  multiplied  by  9,  the  units 
of  the  multiplier,  and  the  product  set  down,  as  in  the  preceding  ex- 
amples. The  multiplicand  is  then  multiplied  by  2,  the  tens  of  the 
muUiplier,  as  before,  the  first  figure  of  the  product  is  placed  under 
the  2,  in  the  place  of  tens.  The  t\¥o  products  are  then  added,  and 
their  sum  is  the  whole  product  or  answer. 

Examples. 

f.  2.  3.         .  4. 

37934      769308      4980076      763896 
2  3  4  5 


Prod.  75868. 


SIMPLE  MULTIPLICATION. 


29 


5. 
67589 
6 

6. 
503'; 

'64      : 

7 

7.          8. 
3818295      9164785 
8           9 

Prod.  405534 

9. 
4879567 
10 

10. 

5864794 
11 

11. 

€583478646 
12 

Prod. 

64512734 

12. 

6357534 
47 

13. 

8324629 
59 

14. 

46293845 
106 

44502738 
25430136 

277763070 
46293845 

Prod.  298804098 

4907147570 

15. 

647906 
4873 

16. 
760483 
9152 

17. 

91867584 
6875 

^157245938 

18.  Multiply  103  by  sixty  seven.     Ans.  6901. 

19.  Said  Jack  to  Harry,  you  have  only  77  chesnuts,  but  I  have 
seven  times  as  raany  ;  how  many  have  I  ?     Ans.  539. 

20.  If  four  bushels  of  wheat  make  a  barrel  of  6our,  and  the 
price  of  wheat  be  one  dollar  a  bushel,  what  will  225  barrels  of 
Hour  cost  ?     Ans.  900  dolls. 

21.  Eighty  nine  men  shared  equally  in  a  prize,  and  received  17 
dolls,  each  ;  how  much  was  the  prize  ?     Ans.  1513  dolls. 

22.  Multiply  308879  by  twenty  thousand  five  hundred  and  three. 
Ans.  6332946137. 

In  some  cases  the  operations  of  multiplication  are  shortened  by 
particular  rules.     Several  Cases  follow. 

Note.  A  composite  number  is  the  product  of  two  or  more  num- 
bers, as  27,  which  3x9,  and,  as  315,  which  =?  5X7X9. 

CASE  I. 
When  the  multiplier  is  a  composite  number,  multiply  the  multi- 
plicand by  one  of  those  figures,  first,  and  that  product  by  the  other, 
kc.  and  the  last  product  will  be  the  total  required.* 

*  The  rcafon  of  this  method  is  obvious  :  For  any  number, multiplied  by  the 
component  parts  of  another  number,  muft  give  the  fame  prodiitft,  as  though  it 
were  multiplied  by  that  number  at  once:  Thus,  in  example  firft,  5  times  the 
product  of  7,  multiplied  into  the  given  number,  makea  33  times  that  given 
number,  as  plainly  as  5  times  7  makes  35- 


m 


SIMPLE  MULTIPLICATION. 


1. 
Mult.  69375  by  35. 
7 

7x5  =  35 

415625 
5 


2078125 


Examples, 
2. 
39187  by  48. 


3. 

91632  by  56- 


4. 
3065  by  63. 
6. 
14567  by  144. 


6. 
6061  by  12K 


CASE  n. 

When  there  are  cyphers  on  the  right  hand  of  either  the  multipfteand, 
or  multiplier,  or  both  :  Neglect  those  cyphers  ;  then  place  the  sig- 
Dificant  figures  under  one  another,  and  multiply  by  them  only  ;  add 
them  together,  as  before  directed,  and  place  to  the  right  hand  as 
many  cyphers  as  there  are  in  both  the  factors. 

Examples. 
1.  2.  3. 

67910  956700  930137000 

5600  320  9500 


Prod.      380296000         306144000         8836301500000 


CASE  in. 

To  multiply  by  10,  100,  1000,  «^c. :  Set  down  the  multiplicand  uq- 
derneath,  and  join  the  cyphers  in  your  multiplier  to  the  right  hand 
of  them.* 

Examples. 
1.  2.  3.  4. 

57935  84935  613975  8473965 

10  100  1000  10000 


Prod.    679350 


CASE  l\^. 

7b  multiply  by  9^9,  999,  «^'c,  in  one  line  ;  Place  as  many  dots  at 
the  right  hand  of  the  multiplicand,  as  there  are  figures  of  9  in  your 
multiplier,  which  dots  suppose  to  be  cyphers  ;  then,  beginning  with 
the  right  hand  dot,  subtract  the  given  multiplicand  from  the  new  ane, 
and  the  remainder  will  be  the  product. t 

*  This  is  evident  from  the  nature  of  numbers,  fince  every  cypher  annexed  to 
:iic  right  of  a  number  incrcafes  the  value  of  that  number  in  a  tenfold  proportion. 

f  Here  It  may  eafily  ht  fcen  that, if  you  multiply  any  fum  by  9,  the  produift 
will  be  hut  9  tenths  of  the  producSk  of  the  fame  fum,  multiplied  by  10  :  and  as 
the  annexing  of  a  clot  or  cypher,  to  the  right  hand  of  the  multiplicand,  fuppof- 
es  it  to  be  increafed  tenfold  :  therefore,  fubtradling  the  given  multiplicand 
from  the  tenfold  multiplicand,  it  is  evident  that  the  remainder  will  be  ninefold 
jhe  faid  given  multiplicand,  equal  to  the  produ^b  of  the  fame  by  9;  and  the 
*am.c  will  hold  true  of  any  number  of  nines. 


SiMPLE  MULTIPLICATION. 


31 


Examples. 

1. 

2. 

3. 

6473.. 

867389 . . . 

5384976...  ^ 

99 

999 

9999 

640827 


53844376024 


That  these  examples  may  appear  as  clear  as  possible,  I  will  il« 
lustrate  them  by  giving  another. 


Mult.  371967.., 
by         999 

371595033 


C  According  to  the  rule,   ) 
(      it  will  stand  thus.        ^ 


371967...  Minuend. 
371967  Subtrah. 


371696033  Rem.   or 
total  Pro. 


CASE  V. 

To  multiply  by  13,  14,  15^  4-c.  to  19;  also  from  101  to  109,  from 
iOOl  to  1009,  ^c. :  Multiply  with  the  unit  figure  only,  of  the  mul- 
tiplier, removing  the  product  one  place  to  the  right  hand  of  the 
multiplicand,  and  so  many  places  further  as  there  may  be  cyphers 
between  the  significant  figures  ;  then  add  all  together,  and  their 
sum  will  be  the  product.* 


Examples. 

1. 

2. 

3. 

75964X13 

7598x104 

G735X100. 

227892 

30392 

33675 

Prod.      987532 


CASE  VI. 

To  multiply  by  21,  31,  41,  4'c.  to  91,  also  by  the  same  figures  with 
any  number  of  cyphers  betzeeen  them  :     Multiply  by  the  left  hand  fig- 
ure, only,  of  the  multiplier,  and  set  the  unit  figure  of  the  product 
one  place  to  the  left,  and  as  many  places  further  as  there  are  cy 
phers  between  the  significant  figures  ;  and  add  the  numbers  togeth 
er  for  the  product. 

Examples. 
1.  2.  3, 

73918  X21  66934  X301  45936  X400i 

147836  170802 


Prod.   1552278 


17137134 


4. 
3167X500001. 


*  The  reafon  of  this  Rule,  and  of  the  following  one  alfo,  will  be  cvideut  or 
infpedking  an  example  under  each  rule, 


32  DIVISION, 


CASE.  VII, 


To  multiply  any  number^  by  any  number^  giving  only  the  Product  ! 
Put  down  the  product  figure  of  the  first  figure  of  the  multiplicand 
by  thej^r^^  of  the  multiplier.  To  the  product  of  the  second  of  the 
multiplicand  by  the  Jirst  of  the  multiplier,  add  the  number  to  be 
carried,  and  the  product  of  the^;*si  of  the  multiplicand  by  the  second 
of  the  multiplier  v  then,  carrying  for  the  tens  in  the  sam,  put  down 
the  rest.  To  the  product  of  the  third  of  the  multiplicand  by  the 
first  of  the  multiplier,  add  the  number  to  be  carried,  and  the  pro- 
duct of  the  second  of  the  multiplicand  by  the  second  o(  the  multipli- 
er, also  the  product  of  the  first  of  the  multiplicand  by  the  third  of 
tlie  multiplier,  carry  the  tens,  and  put  down  the  rest,  and  so  pro- 
ceed till  you  have  multiplied  the  highest  of  the  multiplicand  by  the 
lorn-est  of  the  multiplier.  Then  multiply  the  highest  of  the  multi- 
plicand by  the  second  of  the  multiplier  :  Add  the  number  to  be 
carried,  and  the  product  of  the  last  but  one  of  the  multiplicand  by 
the  third  of  the  multiplier,  and  the  product  of  the  last  but  two  of 
the  multiplicand  by  the  fourth  of  the  multiplier,  &c.  Then  to  the 
product  of  the  last  but  one  of  the  multiplicand  by  the  fourth  of  the 
multiplier  ;  and  so  proceed  till  you  have  multiplied  the  last  of  the 
multiplicand  by  the  last  of  the  multiplier,  which  finishes  the  work. 

Example.  Explanation,       ,    , 

Mult.  632141S  — 

By  2354  5x4—20 

Prod.  12526610910  1X44-24-5X5=31 


4x4-f3-r  1X54-5X3=39 

1x44-3+4x5+1x34-5x2=40 

2x4+4+1x5+4x3+1x2=31 

3x4+3+2x5+1x3+4x2=36 

5x4+3+3x5+2x3+  1x2=46 

6X  5+4+3X  3+2X  2=42 

5x  3+4+3X    2=25 

'.  5x2+2=12 


DIVISION 

TEACHES  to  separate  any  number  or  quantity  given,  into  any 
number  of  parts  assigned  ;  or  to  find  how  often  one  number  is  con- 
tained in  another ;  and  is  a  concise  way  of  performing  several  Sub- 
tractions. 


SIMPLE  DIVISION.  33 

There  are  four  parts  to  be  noticed  in  Division,  \u. 

The  Dividend,  is  the  number  given  to  be  divided. 

The  Divisor^  ii  the  nuoiber  given  to  divide   by. 

The  Quotient,  or  answer  to  the  question,  shows  how  many 
times  the  divisor  is  contained  in  the  dividend. 

If  there  be  any  thing  left  after  the  operation  is  performed,  it  is 
called  the  Remainder  ;  sometimes  there  is  a  remainder  and  some- 
limes  there  is  not.  The  remainder,  when  there  is  any,  is  of  the 
same  denomination  as  the  dividend. 

Division  is  both  Simple  and  Compound. 

Proof. 
Multiply  the  divisor  and  quotient  tog^elher,  and  add  the  remain- 
der, if  there  be  any,  to  the  product ;  if  the  work  be  right,  that  sum 
will  be  equal  to  the  dividend. 

SIMPLE  DIVISION 

Is  the  dividinj^  of  one  number  by  another,  without  regard  to 
(heir  values:  As  56,  divided  by  8,  produces  7  in  the  quotient: 
That  is,  8  is  contained  7  times  in  56.* 

Rule. 

Having  drawn  a  curve  line  on  each  side  of  the  dividend  and 
placed  the  divisor  on  the  left  hand  of  it, 

Seek  how  many  times  the  divisor  is  contained  in  the  least  num- 
ber of  the  figures  of  the  dividend  on  the  left  hand  that  do  contain 


♦  According  to  the  rule,  we  refolv^c  the  dividend  Into  parts,  and  find,  by  tri- 
al, the  number  of  times  the  divlfor  Is  contained  In  each  of  thofc  parts ;  and  the 
only  thing  which  remains  to  be  proved, Is,  that  thefeveral  figures  of  the  quo- 
tient, taken  as  one  number,  accordlno  to  the  order,  in  which  they  arc  placed, 
are  the  true  quotient  of  the  whole  dividend  by  the  dlvifor ;  which  may  be  thus 
dcmonftrated. 

Dent.  The  complete  value  of  the  firft  part  of  the  dividend,  is,  by  the  nature 
of  notation,  10,  101',  1000,  &c.  times  the  fimple  value  of  what  is  taken  in  the' 
operation;  accordingly,  as  there  are  1,  2,  or  o,  &c.  figures  ftandlng  before  it  ; 
and,  confequently,  the  true  value  of  the  quotient  figure,  belonging  to  that  part 
of  the  dividend,  is  alfo  iO,  100,  1000,  &c.  times  its  fimple  value  ;  but  the  true 
Value  of  the  quotient  figure,  belonging  to  that  part  of  the  dividend,  found  by  the 
rule,  is  alfo  U»,  iOO,  10(X),  &c.  times  Its  fimple  value;  for  there  are  as  many  fig- 
ures f<?t  before  it,  as  tlit  number  of  remaining  figures  in  the  dividend  :  therefore 
the  firft  quotient  figure,  taken  in  its  complete  value  from  the  place  it  ftands  in, 
is  the  true  quotient  of  the  dlvifor  in  ilie  complete  value  of  the  firft  part  of  the 
dividen  J.  For  the  fame  reafon,  all  the  reft  of  the  figures  of  the  quotient,  taken 
according  to  their  places,  are,  each,  the  true  quotient  of  the  dlvifor,  in  the  com- 
plete value  of  the  leveral  parts  of  the  dividend  belonging  to  each  ;  becaufe,  as 
the  firft  figure,  on  the  right  hand  of  each  fucceedlng  part  of  the  dividend,  has 
a  lefs  number  of  figures  ftandlng  before  it,  fo  ought  their  quotients  to  have ;  and 
fo  they  arc  adtually  ordered;  confequently,  taking  all  the  quotient  figures  in 
order,  as  they  afc  placed  by  the  rule,  they  make  one  number,  which  is  equal  to 
the  fum  of  the  fue  quotients  of  all  the  feveral  parts  of  the  dividend  ;  and  is, 
therefore,  the  true  quotient  of  the  whole  dividend  by  the  divlfor. 

That  no  obfcurity  may  remain,  in  the  dciTienftration,  it  i$  i!luftrared  by  the 
following  txaTOfile, 


34  SIMPLE  DIVISION. 

it,  and  place  the  answer  on  llie  right  of  the  dividend  ior  the  quo 
tient ;  multiply  the  divisor  bj  this  quotient  figure,  place  the  pro- 
duct under  those  left  hand  figures  of  the  dividend  ;  then  subtract  it 
(herefrom,  and  bring  down  the  next  figure  of  the  dividend  to  the 
right  hand  of  the  remainder  :  If,  when  jou  have  brought  down  a 
figure  to  the  remainder,  it  is  still  less  than  the  divisor,  a  cypher 
must  be  placed  in  the  quotient,  and  another  figure  be  brought 
down  ;  after  which,  you  must  seek,  multiply  and  subtract,  till  yoxi 
have  brought  down  every  figure  of  the  dividend; 

Divlfor  23)74503  Dividend 
ill  part  of  the  dividend  is  =  74000 

26  X 2000  =  50C00  —  —  2000  the  Ifl  quotient. 


1ft  tcmainder  =  24000 
add        500 

•.^d  part  of  the  dividend  =  24500 

25  X  bOO  =  22500 »     900  the  2d  quotient. 

2d  remainder  =    2000 
add         00 

5d  part  of  the  dividend  =  2000 

25  X  bO  =  2000 80  the  2d  quotient 

00 
add        3 

4th  part  of  the  dividend  =         3 

25X0        0 0  the  4th  quotient, 

Laft  rcmaiiider  =  3  —  29i]0  =  Sum  of  all  the  quo- 
tients, or,  the  Anfwer. 

Explan.  It  is  evident  tHi:  dividend  is  refolved  intothefe  parts,  74000-|-500-|- 
00-|-^?  ;  for  the  firft  part  of  the  dividend  is  confidercd  only  as  74;  but  yet  it  is, 
truly,  74000;  and  therefore  its  quotient, inftej«d  of  2,  is  ^OoO, and  the  remainder 
240*  «;  ;  and  fo  of  the  reft  ;  as  may  be  fecn  in  the  operation. 

When  there  is  no  remainder  to  a  divifjon;  the  quotient  is  the  abfolutc  and 
perfetSl  anfwer  to  the  queflion  ;  but  where  there  i&  a  remainder,  it  may  be  ob- 
ferved,  that  it  goes  fo  much  towards  another  time  as  it  approaches  the  divifor; 
thus,  if  the  remainder  be  half  the  divifor,  it  will  go  half  of  a  time  more,  and  fo 
on  ;  in  order,  therefore,  to  complete  the  quotient,  put  the  laft  remainder  to  the 
end  of  it,  above  a  line,  and  the  divifor  below  it.  Hence  the  origin  of  vulgar 
fractions,  which  are  treated  of  hereafter. 

It  is  fomctimes  difficult  to  find  how  often  the  divifor  may  be  had  in  the  num- 
bers of  the  feveral  fteps  of  the  operation  :  The  beft  way  will  be  to  find  how 
often  the  firft  figure  ot  the  divifor  may  be  had  in  the  fit  ft,  or  two  firft  figures  of 
the  dividend,  and  the  anfwer,  made  lefs  by  one  or  two, is,  generally,  the  figure 
wanted  :  but,  if,  after  fubtradling  the  produdl  of  the  divifor  and  quotient  irom 
the  dividend,  the  remainder  bt  equal  to,  or  exceed  the  divifor,  the  quotient 
figure  muft  be  increafed  accordingly  ;  or,  if  the  product  of  the  divifor  and  quo- 
tient figure  exceed  tlic  dividend,  then  the  quotient  figure  muft  be  proportiona- 
bly  lefiened. 

riie  Tcafon  of  the  method  ol  proof  is  plain;  for,  fiilce  the  quotient  is  the 
number  of  times  the  dividend  contains  the  diviTor,  the  product  of  the  quotient 
and  d»vifor,muft,  evidently,  be  equal  to  the  dividend. 


SIMPLE  DIVISION. 

Examples. 


qo 


pivifqr.  Dividend.  Quotient. 
3)175817(68005 
15 


^34 


18 

18 

17 

15 

2  Rem. 

Proof.               ^ 

.*8G05  Q^'iotient, 

X3  Divisor  -|- 

2 

In  this  example,  I  find  that  S,  the  di- 
visor,  cannot  be  contained  in  the  first  flg- 
iire  of  the  dividend  ;  therefore,  I  take  two 
figures,  viz.  17,  and  inquire  how  often  3 
is  contained  therein,  which  finding  to  be 
5  times,  I  place  the  5  in  the  quotient,  and 
muhiply  the  divisor  by  it, setting  the  first 
figure  of  the  multiphcation  under  the  7 
in  the  dividend,  &c.  I  then  subtract  15 
from  17,  and  find  a  remainder  of  2,  to  the 
right  hand  of  which  I  bring  down  the  next 
figure  of  the  dividend,  viz,  5  ;  then  I  in- 
quire how  often  the  divisor  3,  is  contain- 
ed in  25,  and,  finding  it  to  be  8  times,  I 
multiply  l)y  8,  and  proceed  as  before,  till 
1  bringdown  the  1,  when,  finding  I  can- 
not have  the  divisor  in  1,  I  place  0  in  the 
quotient,  and  bring  down  7  t^  the  t,  and 
proceed  as  at  the  first. 


i75817 

Observe,  that,  in  multiplying  by  3,  I  add  in  the  2 


^. 


29)153598(5296 


Hi 


3. 
6493)91876375(14150 
6493 


85 


26946 
25972 


279 
^61 


188 
174 

14 


9743 
G493 

32507 

32465  . 

425 


There  are  fevcral  other  methods  made  ufc  of  to  prove  divifion;  as  follow,  vi«. 

Rule  I. 

Subtrai^t  the  remainder  from  the  dividend;  divide  this  number  by  the  quo- 
ijent,  and  the  quotient,  found  by  this  divifion,  will  be  equal  to  the  former  divi- 
for,  when  the  work  is  right. 

Rule  II. 

Add  the  remainder  and  all  the  produdls  of  the  feveral  quotient  figures  mul- 
fipUcd  by  the  divifor  together,  according  to  the  order  In  which  they  ftand  in  t  he 
work,  and  the  fum,  when  the  work  is  right,  will  be  equal  to  the  dividend. 

Here  the  nun\bers  to  bp  added  are  the  produ<£ts  of  the  divifor  by  every  fig- 
arc  of  the  quotient,  feparately ;  and  each  by  its  place,  poflefles  its  complete 
value  ;  therefore,  the  fum  of  the  parts  together  with  the  remainder,  muft  be 
f-qual  to  the  whole.  I  will  illuftrate  the  whole  by  an  example  proycd  accord- 
i.ng  to  the  feveral  different  methods. 


36  SIMPLE  DIVISION. 

4.  6.  6. 

28)503775(  35)197184(  «5)994466( 

7.  8.  9. 

236)3798567(      3479)483956796(    5679) 1 9647394 ( 

10.  11. 

38473)119184693(       641976)9187642959r 

12. 

5823789(791 822376496( 

13. 

123456789)121932631112635269( 

Some  operations  are  more  readily  performed  by  the  following^ 
particular  rules. 

CASE  1. 

When  there  is  one  cypher y  or  more,  at  the  right  hand  oj  the  divisor  : 
It  or  they  must  be  cut  oft ;  also  cut  oft'  the  same  number  of  fig- 
ures from  the  dividend,  and  then  proceed  as  in  Case  first :  But  the 
figures  which  were  cut  off  from  the  dividend  must  be  placed  at  the 
right  hand  of  the  remainder.! 

70)9  8  7  6  5  4  3  2  l(125019.'i3 

7  9»  79-f  34  remainder. 

19  7  11^2517577 

15  8*  87513671 

■   +34 

.396 

.395*  987654321  Proof  by  MuItiplJcatioK. 

...    154 

.   .   .    .   7  S»  987654321 

.   .   . _-34 

....    7  5  3 

....    7  1   1*         12501953)987654287(79  Proof  by  Divi%>D. 

.... 87513671 

4  2  2 

3  9  5*  112517577 

112517577 

2  7   1 — 

.   .    ....   2  3  7* 

'.    '.    .    .    .'   .*   .     3  4* 

9  8  7  6  5  4  3  2   1     Proof  by  Addition. 
We  need   only  to  refer  to  the  example,  except    for  the  proof  by  addition  ; 
where  it  nviy  be  remarked,  that  the  Afterifms  fliew  the  numbers  to  be  added, 
and  the  dotted  lines  their  order. 

f  The  reafon  of  this  contradlion  it  is  cafy  to  conceive;  for  cutting  off  the 
fame  figures  from  each,  is  the  fame  as  dividing  each  of  them  by  10,  100,  1000, 
&c.  and  it  is  evident,  that  as  often  as  the  whole  divifor  is  contained  in  the  whole 
dividend,  fo  often  mud  any  part  of  the  divifor  l>e  contained  in  the  like  part  of 
the  dividend;  this  method  is  only  to  avoid  a  needlefs  repetition  of  cyphers-, 
which  would  happen  in  the  common  way,  as  may  be  fecn  by  working  one  of 
the  examples  of  this  cafe  in  the  common  way  without  cutting  off  the  cyphers. 


SIMPLE  DIVISION.  37. 

Examples, 
1.  2. 

65|00)3794326|75(58374  5193|000)8937643|893( 

325 

544  3. 

520  917|0)47658|3( 


243 

195  4. 


87^j00P)91764789430|00Q( 

482 
'455 

276 
260 

1675  Rem. 

5.  6.  7. 

Q.uot.  Rem.       Q,uot.  Rem.         Q,uot.  Rem. 

1|0)9584|(5     '  I|00)76495l80  li0OO)93751339|462 

Note.  In  dividing  by  10,  100,  1000,  &c.  when  you  cut  off  as 
many  figures  from  the  dividend,  as  there  are  cyphers  in  the  divi- 
sor, your  work  is  done  ;  those  figures,  cut  off  at  the  right  hand,, 
aj^e  the  remainder,  and  those  on  the  leA,  the  quotient,  as  above. 

CASE  II. 

Short  Division  may  be  used  when  the  divisor  does  not  e;(ceed  12. 
it  is  performed  by  the  following 

Rule, 
First,  seek  how  often  the  divisor  can  be  had  in  the  first  figure, 
or  figures  of  the  dividend  ;  which,  when  found,  place  in  the  quo- 
tient; then,  mentally,  multiply  your  divisor  by  the  figure  placed  in 
the  quotient,  and  subtract  the  product  from  the  like  number  of  the 
left  hand  figures  of  your  dividend,  and  the  remaining  units,  if  any, 
must  be  accounted  so  many  tens,  which  you  must  suppose  to  stand 
at  the  left  hand  of  the  next  figure  in  the  dividend,  and  to  be  reck- 
oned with  it;  then,  seek  how  often  you  can  have  your  divisor  is 
those  two  figures  ;  but,  if  nothinq^  remain,  you  must  then  seek  how 
often  your  divisor  is  contained  in  the  next  figure,  or  figures,  and 
thus  proceed  till  you  have  done. 

Examples. 
Divisor.  Dividend.         2.  3.  4.  6. 

2)71935  3)51903       5)633795      6)8471937     7)193847 


Quot.  35967—1  rem. 


^  iDlMPLE  DIVISION. 

6.  7.  8.  d. 

3)5437846         9)45963784  11)91843756       12)1196437847536 


CASE  III. 

When  the  divisor  is  such  a  number  thai  any  t7a;o,  or  more,  figures  in 
zhe  Table,  being  multiplied  together,  wiU  produce  it:  Divide  tire  giv- 
en dividend  by  one  of  those  figures;  the  quotient,  thence  arising, 
by  (be  other,  and  so  on  ;  and  the  last  quotient  will  be  the  answer.'^ 

♦  This  follows  from  the  contradtion  in  cafe  3d, of  Simple  Multiplication, of 
which  it  Is  only  the  rcverfe  ;  for  the  fourth  part  of  the  half  of  any  thing  is  evi- 
dently the  fame  as  the  eighth  part  of  the  whole  ;  and  fo  of  any  other  number. 

As  the  learner  at  prcfcnt  is  fuppofed  to  be  unacquainted  with  the  nature  of 
fradlions,  and  as  the  quotient  is  incomplete  without  the  remainder  ;  I  fliall  here 
give  a  rule  for  finding  the  true  remainder,  without  having  recourfc  to  fratftions. 

Rule  I. 

JifuUiply  the  quotient  by  the  divifor  :  Subtradl  the  product  from  the  divi- 
dend and  therefult  will  be  the  true  remainder. 

The  Rule,  which  is  moft  commonly  made  ufc  of,  when  tbc  divifor  is  a  com* 
jjoiite  number,  is 

Rule  II. 

Multiply  thelaft  remainder  by  the  preceding  divifor,  or  laft  but  one,  and  to 
the  produ<5t  add  the  preceding  remainder  ;  multiply  this  fum  by  the  next  pre- 
ceding divifor,  and  to  the  product  add  the  next  preceding  remainder ;  and  fc 
on,  till  you  have  gone  through  all  the  diviforsand  rcniainders,to  the  firft. 

Example. 

6)35397  divided  by  15,Q  1  the  laft  remainder. 

— multiply  by  ♦,')  the  laft  divifor  but  one, 

5)14'2r>2— 5  — 

.')'2^46— -2  add  2  the  fecond  remainder, 

560—1  7 

multiply  by  6  the  firft  divifor. 

Ans.  509-i-^;V  42 

add     5  the  firft  remainder. 

47  the  true  remainder, 
To  explain  this  rule  from  the  example,  we  may  obfervc,  that  every  unit  in  the 
L'rft  quotient  may  be  looked  upon  as  containing  6  of  the  units  In  the  given  divi- 
ilcnd;  conferuently,  every  unit  which  remains,  will  contain  the  fame;  there- 
i^ore,  tbii  remainder  muft  be  multiplied  by  6,  to  find  the  units  it  contains  of  the 
j^lvf  n  dividend.  Again,  each  unit  in  the  next  quotient  will  contain  5  of  the  pre- 
i'cding  ones,  or  30  of  the  firfl,  that  is,  6  times  5 ;  therefore,  what  remains  muft  b^ 
iTtultiplied  by  30,  or,  which  is  tlie  fame  thing,  by  G  and  5  continually :  Now, 
this  is  the  fame  as  the  Rule  ;  for  inftead  of  finding  the  remainders,  fcparately , 
they  are  reduced  from  the  bottom,  upwards,  ftcp  by  ftep,  to  one  another,  and 
che  remaining  units,  of  the  fame  clals,  taken  as  they  occur. 


iJiMPLE  DIVISION. 

1st.  method, 
9)196473 

8)21830 

^uoU  2728— 

■57 

Examples, 
2c?.  method, 
8)196473 

9)24559 

Qnot.  2728-^57 

3f/.  methods 
72)196473(2728  Qiwt 
144 

524 
604 

207 
1^4 

633 
676 

57  RtimaintJer. 

I  have  wrought  the  above  question  three  way?,  that  the  learner 
may  understand  the  method  of  finding  the  true  re.naindcr,  accord- 
ing to  this  case.  In  the ^trsr,  in  dividing  by  9,  3  remains,  and  by  8, 
6  remains;  which  being  the  last  remainder,  1  mtiUiply  it  by  the 
first  Jivisor  9,  and  add  in  the  first  remainder  3,  and  ihey  make  57, 
the  true  remainder.  In  the  second  method,  dividing  by  8,  1  re- 
mains, and  by  9,  7  remains  ;  I  therefore,  multiply  7,  the  last  re- 
mainder, by  8,  adding  in  the  1,  and  they  make  57  as  before.  The 
third  method  is  self  evident,  and  shews  that  the  other  remainders 
are  true. 

2.                             3.                           4.  5. 

36)79638  25)197835             87)93975  54)93738764 

6.                                     7.  8. 

121)75323939                  132)38473692  ,  144)891376429732 

CASE.  IV. 

IVIien  (he  divisor  is  a  whx)le  nvtnbcr  zvith  some  part  of  unity,  as  S-^, 
5},  6|,  &c.  proceed  by  either  of  the  tbllowing  methods. 

1.  Multiply  the  whole  ntimber  in  the  divisor  by  the  number  of 
parts  into  which  unity  is  divided  in  the  fraction,  and  to  the  product 
add  the  number  of  parts  of  unity  taken  in  the  fraction,  and  the  di- 
visor will  be  reduced  to  the  parts  indicated  by  the  fraction,  for  a 
new  divisor.  Multiply  the  dividend  by  the  parts  into  which  unity 
is  divided  in  the  fraction,  for  a  ne^v  dividend.  Divide  the  new  div- 
idend by  the  new  divisor,  and  the  quotient  will  be  the  answer. 

Ex.  1.   Divide  1820  by  21. 

Here,  I  multiply  2  by  3,  and  add  1  to  the  product,  and  have  7, 
for  the  new  divisor.  Then  I  multiply  1820  also  by  3,  and  have 
5460  for  the  new  dividend.  Then  5460-7-7=780,  the  answer,  or 
21  is  contained  in  1820,  exactly  780  times. 

jVote.  It  is  obvious  that  the  dividend  and  divisor  are  propor- 
tionally increased  by  the  multiplication,  so  that  the  quotient  wil? 
be  the  same  as  if  they  had  not  been  thus  increased. 

Ex.  2.   Divide  G375  by  5^ . 


40  SIMPLE  DIVISION. 

Proceediog  as  before,  the  new  divisor  k  J  I,  and  the  new  divi- 
dend is  12760.  Then  12760—11=1159  and  1  remains,  or  1169^^- 
ihe  number  of  times  5i  is  contained  in  6375. 

3.  Divide  10142  by  3-|.     Ans.  2766. 

4.  Divide  170  by  2f . 

6.   Divide  168765  by  15|. 

II.  Proceed  according  to  the  general  rule  for  division,  being 
careful  to  add  the  value  of  the  fractions  to  the  several  remainders, 
at  every  step  of  the  process. 

Ex.  1.  Divide  1820  by  2^. 
2^)1820(78(5 

Here  7x2i=16|,  which  being  subtracted  from 
18,  leaves  1|.  [Now  f  belongs  to  the  place  of 
hundredths,  and  is  f  of  10  for  the  place  of  tenths. 
But  I  of  10=6|,  to  which  add  Ihe  2,  and  we  have 
8|to  be  annexed  to  the  1  remainder,  and  we  have 
18|  for  the  true  remainder.     Now  2}  is  contain- 

ed  in  18|  exactly  8  times.     The  rest  of  the  pro- 

0  cess  is  evident. 

Ex.  2.  Divide  6375  by  6^. 
51)6375(11 69 Jy 

—  Here  once  6|-  taken  Irom  6,  leaves  a.     But 

|-  l  in  thousandths  place  is  5  in  the  place  of  hun- 

—  dredlhs,  and  5  added  to  the  3  hundredths  is  8. 
8  Taking  from  8,  once  5^,  2i  remain.  As 
5  J  before,  the  i  becomes  5  in  the  place  of  tenths, 

—  whicJh  added  to  7  tenths,  make  12,  which 
2|  added  to   the  remainder  2  in   the  preceding 

' place,  give  32.     From  this  6  times  5}  are  to 

32  be  taken,  and  4i  remain.     But  4^  in  the  place 

-7]^  of  tenths,  is  45,  which  added  to  5  make  50. 

— -  From  50  take  9  times  5i,  and  i  remains.    Now 

4i  in  5i  are  11  halves,  so  that  ^  is  ^\,  which  an 

nexed  to  the  quotient,  gives  the  whole  quo- 

60  tient. 
49i 


Ex.  3.  Divide  347  by  2|     Ans  144-i%. 
4.  Divide  13567  by  13f. 

Supplement  to  Contractions  in  Multiplication. 
1.  The  shortest  method  of  multiplication,  when  the  multiplier  is 
any  even  part  of  100,  1000,  &c.  is  by  division  :  For  if  the  mnliipli- 
candbe  increased  by  a  number  of  cyphers  equal  to  the  number  of 
places  in  the  multiplier,  and  a  part  of  that  product  taken  for  the 
same  proportion,  which  the  multiplier  bears  to  1,  an<l  the  same 
number  ot  cyphers  annexed  to  it,  the  quotient  will  be  the  true  pro- 
duct. 


SIMPLE  DIVISION.                                41 

1.  Multiply  39756  into  125.  2.  Multiply  57638  by  33i 

125=1  of  1000,  wherefore^  33i=i  of  100,  therefore, 

8)39756000  3)5763800 

4969500  Product.  1921266f  Product. 


3.  Multiply  91378  by  3331.      Aos.  304693331. 

The  reason  of  the  preceding  rule  is  obvious. 

U.  If  any  digit,  with  cyphers  annexed,  be  divided  by  9,  the  quo- 
tient will  consist,  wholly,  ot  such  digits,  and  so  many  9ths  of  an 
unit  over;  hence  the  following  method  of  multiplying  by  repe- 
tends  of  any  of  the  digits. 

The  reason  of  the  method  may  be  seen  by  the  first  example. 
Thus  80000—9,  or  ^^f^^=8888f,.  and  by  multiplying  by  80000 
and  dividing  by  nine,  you  get  eight  9ths,  of  645  too  much,  which 
must  of  course  be  subtracted  to  obtain  the  true  product.  Or,  thus, 
8oio£_i^388g  .  then  645x8888=645x^^^^-— 645xf =6733333f 
— 573f =5732760.  But  as  three  9lhs,  belong  to  both  numbers,  and, 
as  an  equal  number  of  9ths,  will  belong  to  both  numbers  in  any 
case,  it  is  not  necessary  to  write  them,  as  they  balance  each  other 
in  the  subtraction. 


1. 

645  by  8888 
80000 

2. 
5394  by  66666. 
600000 

3. 

3798  by  444 

Prod.  1686312 

9)51600000 

6733333 
Subtract     573 

9)3236400000 

369600000 
Subt.       3696 

Product.  5732760 

Prod.   359596404 

• 

III.  When  the  multiplicand  has  a  fraction  belonging  to  it,  such 
as  one  fourth,  one  half,  &c.  add  such  a  part  of  the  multiplier  as 
the  fraction  signifies,  to  the  product  obtained  by  multiplying,  and 
the  sum  is  the  whole  products  When  such  a  fraction  belongs  to 
the  multiplier,  add  to  the  product  such  a  part  of  the  multiplicand 
as  the  fraction  denotes. 

1.  Multiply  153-1  by  6, 
6 


918  Product. 
3  half  of  6. 


921  Answer. 

It  is  evident  that  6  halves  or  3,  must  he  added  to  153x6  to  ob- 
tain 6  times  153i.  And  the  same  must  be  true  in  every  similar 
case. 

F 


42  SIMPLE  DIVISION. 

2.  Multiply  638  by  6f 


3828   Product  by  6. 
212|  i  of  the  multiplicanJ. 


4040|  Ans. 
As  the  multiplier  is  6|,  it  i^  evident,  that  to  6  times  638,  there 
must  be  added  one  third  of  638,  the  multiplicand,  for  the  whole 
product.     'J'he  third  part  of  638  is  evidently  21^2,  and  2  remain- 
der, or  |,  because  it  is*  2  parts  of  the  3  in  the  divisor. 

J\''ote.  If  the  sum  of  the  products,  or  quotients,  of  two  or  more 
numbers  multiplied  or  divided  by  the  same  quantity,  be  reqtiired  ; 
then  multiply  or  divide  the  sum  of  the  numbers  by  the  common 
multiplier  or  divii-or,  and  the  product  or  quotient  will  be  the  an- 
swer. If  the  difference  o<  the  products  or  quotients  be  required, 
then  multiply  or  divide  the  dilference  of  the  numbers,  as  before^ 
for  the  answer. 

1.  Required  the  sum  and  difference  of  the  products  of  27  and 
22,  multiplied  each  by  3. 

Now  27x3+22x3  =  the  sum  ==  147=3x49=3x27+22. 

And    27X3—22X3  =  the  diff.  =  16=3x5=3x27—22. 
2    Required  the  sum  and  difference  of  the  quotients,  by  dividing* 
68  and  44,  each  by  4. 

Now  68~4+44-~4  =  the  sum  =  28=112-~.4=68+44-r-4. 
And  68—4—44—4  =  the  diff.  =  6=24—4=68-44-4-4. 
The  same  course  may  obviously  be  pursued  in  any  similar  case. 

Examples  for  Practice. 

1.  Divide  1292  dolls,  equally  amon^  17  men.    Ans.  76  dolls.  eacb> 

2.  Divide  2625  dolls,  among  35  soldiers.     Ans.  73  dolls,  each. 

3.  If  360  cents  are  to  be  divided  equally  among  eighteen  poor 
persons,  how  much  vviil  each  have  ?  Ans.  20  cents. 

4.  If  a  field  of  19  acres  produce  513  bushels  of  wheat,  liow 
much  is  that  for  one  third  of  an  acre  ?  Ans   9  bushels. 

5.  A  man  receives  1095  pounds  a  year,  what  is  it  for  a  day  ? 

Ans.  3  pounds. 

6.  There  are  5^  yds.  in  a  rod  ;  how  many  yards  are  there  in 
40  rods  ?  Ans.  220  yds. 

7.  What  number  must  you  multiply  by  47,  to  produce  298804098  ? 

Ans.  6357534. 

8.  What  number  must  be  multiplied  by  379  to  produce  33789678  ? 

Ans. 

9.  In  a  rod  are  16i  feet  ;  how  many  feet  in  320  rods  ? 

10.  If  3650  pounds  of  bread  are  to  be  divided  equally  among 
365  soldiers  for  10  days,  how  much  will  each  receive  a  day  ? 

Ans.  1  pound, 
n.  Multiply  34678  by  250.  Ans.  866950- 

12.  Multiply  125  by  77777.  Ans, 


TABLES  IN  COMPOUND  ADDITION.  43 


TABLES  IN  COMPOUND  ADDITION. 

1.  Federal  Money.* 

marked.       mills. 
10  Mills     ^  g  rCent  tn.c.  ^        10=       1  cent. 
10  Cents    f  ®    iDime     d.  f       100=     10=     1  dime. 


10  Dimes   (^  jDollar    g.  (     1000=   100:=   10=   1  dollar. 
10  Dollars)  g  (Eagle    E.)  10000=1000=100=10=1  eagle, 

2.  English  Money. 


marked 
4  Farthings  ^  C  Penny         grs.  d. 

12  Pence        >  make  one  <  Shilling       s. 
20  Shillings   )      '  ( Pound.        £. 


*  The  following  account  is  abftractcd  from  the  "  Act  cflablifliing  a  Mint, 
and  regulating  the  Coins  of  the  United  States,"  paflcd  April  and,  1792. 

The  money  of  account  of  the  United  States  (hall  he  cxprcflcd  in  dollars  or 
units,  difmcs  or  tenths  of  a  dollar,  cents  or  hundredths  of  a  dollar,  and  mills  or 
thoufandths  of  a  dollar. 

The  coins  of  gold,  silver,  and  copper  of  the  U.  S.  fliall  he  of  the  following  de- 
nominations, viz. 

.  r    1.  Eagle,  of  the  value  of  ten,  dollars. 

'3  ■<    2.  Half  Eagle,;... .Jive  dolls. 

O   C     -'.   QvARTZk  EAGLZy..........tzvo  and  a  ha// do\h. 

4.  Dollar,  of  the  value  of  the  Spanjb  w///(ri  dollar. 

>.  Half  Dollar,..m... ». ...Aa^the  dollar. 

>  i    f).  Quarter  DoLLARy«.........~....<»n<ryottrM  the  dollar. 

i^   j    T.  Dimes, - ~ «..M«.enf /<r«*i>  of  the  doll. 

B.  Half  Dime,.- ~~.~ one  ttvent'utb  of  the  doll. 

CUj^    C    .9.   Cent, .......ik.......r,...one  hundredth  ol  X\\t  Ao^.. 

y  c.  c  lO.  Half  Cent, -...-.«. one  half  the  cent. 

The  Jlandard  for  all  gold  coins  of  the  U.  S  fliall  be  eleven  parts  of  pure  gold 
and  one  part  of  alloy  in  tivelve  pirts  of  the  coin.  The  alloy  is  to  be  filvcr  and 
copper,  but  the  filvcr  is  not  to  exceed  one  half  in  the  alloy. 

The  Eagle  fliall  contain  two  hundred  ahd  forty  feven  and  a  half  grains  of 
pure  gold, or  two  hundred  and  feventy  grains  of  ftandard  gold;  and  the  other 
gold  coins  in  the  fame  proportion. 

The  Jiandard  for  all  filvcr  coins  of  the  U.  S.  fliall  be  one  thoufand  four  hun- 
dred and  eighty  five  parts  of  pure  filvcr  and  one  hundred  and  fcvcnty  nine  parts 
alloy;  and  tne  alloy  fliall  be  pure  copper. 

The  Dollar  fliall  contain  three  hundred  and  feventy  one  and  one  fourth  grains 
of  pure  filver,  or  of  four  hundred  and  fixteen  grains  of  ftandard  filver;  and  the 
other  filver  coins  in  the  fame  proportion. 

The  Copper  coins  are  to  be  pure  copper.  The  Cent  is  to  contain  eleven  p  cn- 
ny  weights  of  copper;  and  the  Half  Cent  in  proportion. 

The  proportional  value  of  gold  to  filver  in  all  coins  current  by  law  In  the  U. 
S.  fliall  be  fifteen  to  one,  or  fifteen  pounds  weight  of  pure  filver  fliall  be  equal  . 
to  one  pound  weight  of  pure  gold. 

All  coins  of  gold  and  filver,  ilfued  from  the  Mint  of  the  U.  S.  fliall  be  a  law- 
ful tender  in  all  payments  at  the  preceding  values  when  of  full  weight, and  If 
Rot  of  full  weight,  of  proportional  values. 


44 


TABLES  IN  COMPOUND  ADDITION. 


Farthings. 


4  =       1  Penny. 

48  =     12  =     1  Shilling. 

960  =  240  =  20  =  1  Pound. 

A  groat  is  4c?. 

FfiNCE  Tables. 

cl.            S. 

d.      d.          s.      d.               s.           d. 

s.          d. 

20  =  1 

8 

120  =  10     0 

1  =     12 

11  =  132 

30  =  2 

6 

130  =  10  10  ^ 

2  =     24 

12^  =  144 

40  =  3 

4 

140  =  11     8 

3=    36 

13  =  166 

50  =  4 

2 

150  =  12     6 

4  =     48 

14  =  168 

60  =  ^ 

0 

160  =  13     4 

6  =     60 

15  =  180 

70  =  5 

10 

170  =  14     2 

6  ==     72 

16  =  192 

80  =  6 

8 

180  =  15     0 

7  =     84 

17  =  204 

90  =  7 

6 

190  =  15,10 

8  =     96 

18  =  216 

100  =  8 

4 

200  =  16     8 

9  =  108 

19  =  228 

no  ^  9 

o 

240  =  20     0 

10  =  120 

20  =  240 

3.  Troy  Weight.* 

24  Grains 

make  one     Pennyweight,     marked  grs.  put 

20  Pcnnyweig 

hts          -         -        Ounce, 

oz. 

12  Ounces 

- 

=.         -             Poi] 

nd,     - 

^     m 

\G  Drams 
16  Ounces 
28  Pounds    - 
4  Quarters     ~ 
20  Hundred  wt 


Grains. 

24  =       1  Pennyvveighto 
480  =    20  =     1  Ounce. 
5760  =  240  =  12  =  1  Pound. 

4.  Avoirdupois  Weight.! 

make  1     Ounce,  marked  dr.     oz. 

Pound,         -----       iti 

-    Quarter  of  a  hundred  wt,  -  qr. 

Hundred  wt.  or  112  pounds,         -     Cwt. 

-         -     Ton, T. 


*  By  this  weight  are  weighed  Gold,  Silver,  Jewels,  Eledluaries,  and  all  liquors. 

All  ounce  of  gold  is  divided  into  24.  parts,  called  carats^  and  an  ounce  of  fil- 
vcr,  into  20  parts,  called  pennyweights;  therefore,  to  diftinguiili  finenefs  of 
metals,  inch  gold  as  will  abide  the  fire  without  lofs,  is  accounted  24  carats  fine  -. 
If  it  lofe  %  carats  in  triiil,  it  is  tailed  az  carats  fine,  &c. 

A  pound  of  filver  which, lofes  nothing  in  trial,  is  iz  ounces  fine;  but,  if  it 
Jofe  3  pennyweights,  it  is  ii  oz.  17  pwts.  fine,  &c. 

Alloy  is  fome  bafe  metal  with  which  gold  or  filver  is  mixed  to  abate  its  fine- 
nefs ;  22  car-its  of  gold,  and  2  carats  of  copper,  are  cftcemed  the  true  ftandard 
for  gold  coin  in  England,  the  alloy  being  one  eleventh  part  of  the  fine  gold  : 
and  n  oz  2  pwts.  of  fine  filver,  melted  vt^ith  18  pwts. of  copper,  make  the  true 
flandard  for  filver  coin. 

Note.  175  Troy  ounces,  are  precifely  equal  to  192  Avoirdupois  ounces,  and 
175  Troy  pounds  are  equal  to  144  Avoirdupois,  x  lb.  Troy  =:  5760  grain?, 
and  I  lb.  Avoirdupois  =  70CO  grains. 

•{•  By  Avoirdupois  arc  weighed  all  coarfe  and  drofly  goods,  grocery  and  chand- 
lery wares  ;  bread,  and  all  metals,  except  gold  and  filver. 

A  barrel  of  pork  weighs  200  lb.     A  barrel  of  beef,  2Qo  lb.    A  f^nintal  of  fifli. 


TABLES  IN  COMPOU^fD  ADDITIOX. 


45 


Drams. 

16  =  1  Ounce. 

256  =         16  =         1  Pound. 
7168  =      448  =      28  =  1  Quarter. 
28672=     1792=     112=    4=     1  Hund.  wt. 
573440  =  35840  =  2240  =  80  =  20  =  1  Ton. 


5.  Apothecaries'  Weight.* 


20  Grains 
3  Scruples 
8  Drams 

12  Ounces 


make  one 


Grains. 

20  = 

60  = 

480  = 

5760  = 


Scruple, 
Dram,. 
Ounce, 
Pound, 


marked  gr.  9 

-  3 

-  lb. 


1  Scruple. 

3  =     1  Dram. 
24  =    8  =     1  Ounce. 
;88  =  96  =  12  =  1  Pound. 


6.  Cloth  Measure.! 


2  Inches,  and  one  foupth 
4  Nails,  or  9  Inches 

4  Quarters  of  a  yard,  or  36  Inches 

3  Quarters  of  a  yard,  or  27  Inches 

5  Quarters  of  a  yard,  or  45  Inches 

6  Quarters  of  a  yard,  or  54  Inches 

4  Quarters,  1  Inch  &  one  5th,  or 
37  Inches  and  one  fifth 

3  Quarters  and  two  thirds 


make  1    Nail,  marked  in.  na. 
Quarter  of  a  yard,  qr. 


Yard, 
Ell  Flemish, 
Ell  English, 
Ell  French, 

Ell  Scotch, 

Spanish  Var. 


yd. 

E.  Fl. 

E.  E. 

E.  Fr, 

p.  Sc. 


1  Cwt.  Avoirdupois,     ja  particular  things  make  one  dozen  ;  la  dozen  i  grofs, 
and  144  dozen  i  great  grofs.     ao  particular  things  make  1  fcore, 

lb.  A  Stone  of  Iron,  fliot,7  lb. 

56  orhorfeman's  weight,^  I4 

94  Butcher's  Meat,     8 

30  A  gallon  of  Train  Oil       7^ 

2.06  A  Tod  is      -     -     -     -     28 

112  A  Weigh     -     -     -    -    182 

1120  A  Sack      -     -    -     -      364 


A  Firkin  of  Forelgt 
A  Barrel  of — ■ 


^  Punch,  of- 
A  Fother  of- 


|Butter 
%oap 
Anchovies 
Soap 
Raiiins 
Prunes 


Lead       19^  Cwt. 


A  Sack 
Ahft 


4368 


*  All  the  weights  now  ufed  by  Apothecaries,  above  grains,  are  Avoirdupois. 
The  Apothecaries'  pound  and  ounce,  and  the  pound  and  ounce  Troy  are  the 
fame,  only  differently  divided  and  fubdivided. 

t  All  Scotch  and  Irifli  linens  are  bought  by  the  Englifli  or  American  yard, 
which  is  the  fame, and  all  Dutch  linens  by  the  Ell  Fiemifli  ;  but  are  all  fold  in 
America  by  the  American  yard;  though  the  Dutch  liucns  are  fold  in  EugUnc^ 
>y  the  Ell  Englifli,  and  the  Scotch  and  Irifli  linens,  as  in  America. 

The  Scotch  allow  one  Englifli  yard  in  every  fcore  yards. 


4Q 


TABLES  IN  COMPOUND  ADDITION 


Nails,   4  =  1  Quarter. 
16  =  4  =  1  Yard. 
12  =  3  =  1  Flemish  Ell. 
20  =  5  =  1  English  £11. 
24  =  6  =  1  French  Ell. 

7.  Long  Measure.* 

3  Barley  corns  -r         make  1  Inch, 

12  Inches       -         -         .         ^  Foot, 

3  Feet,     -         .         -         -         -  Yard,   - 

51  Yards,  or  16i  feet      -         -  Koc^  Perch,  or  Pole, 

40  Poles Furlong, 

8  Furlongs     -         -         -         .  Mile, 

C91  Statute  miles,  nearly  \  C^^gree  of  a 

^  f  great  Circle, 

A  great  Circle 
of  the  Earth. 
Or^  in  Measuring  Distances. 

make  1  Link. 


marked  bar.  in. 

ft. 

■   yd. 

pol. 

fur. 

mile. 

deg. 


360  Degrees 


7yW  Inches 
25  Links     - 
100  Links 
10  Chains     - 
8  Furlongs 
Bar-  corn?,  3  —  1  Inch 

36  =         12  = 
108  =         36  == 
694  =       198  = 
23760  =     7920  =     660    = 
190080  =  63360  =  6280 
Inches,  1^^-^  =         1  Link, 


Pole. 
Chain. 
Furlong,. 
Mile. 


1  Foot. 
3    = 


161- 


1  Yard. 
61  =      1  Pole. 
220    =    40  =  I  Furlong. 
1760    =  320  =  8  =  1  M. 


198  =      25  =       1  Pole  or  Perch. 

792  =     100  =      4=1  Chain. 

7920  =  1000  =    40  =  10  =  1  Furlong. 

63360  r=:  8000  =  320  =  80  =  8  =  1  Mile. 

8.    TiME.t 

make  1  Minute,  marked  s, 
Houri; 
.     Day,       - 

Week, 
-     Month,     '. 


t!0  Seconds     -         -         -         -         make  J  Minute,  marked  s.  m. 

GO  Minutes HourJ         -         -       h. 

24  Hours       -  -  -  -  -  "     Day,       -  -  d. 

7  Days Week,         -         -    w. 

4  Weeks      .-.-.-     Month,     -  -     mo. 

23  Months,  1  day  and  6  hours       -         -         Julian  year,  yr. 

•  The  ufe  of  l<ong  Meafurc  Is  to  meafure  the  dlftance  of  places,  or  any  other 
iJiing,  where  length  is  confidcrcd  without  regard  to  breadth. 

Note.  6o  geometrical  milts  make  a  degree.  4  inches  a  hand.  5  feet  a 
gcomttrical  pace.  6  points  make  i  line,  12  lines  an  inch,  la  inches  a  foot,  and 
6  feet  one  French  toife,  or  Fathom,  equal  to  6  feet  4  inches,  8,812,875  lines, 
F-n^'Iifli  meafurc.  i  Enslifli  toot  equal  to  ii  inches  31 154  lines  French.  66 
ieet.or  4  poles,  make  a  Gunter's  chain.     3  miles  make  a  league. 

f  By  the  Calendar,  tlie  year  is  divided  in  the  following  manner  : 
Thirty  days  hath  September,  April,  June,  and  November  ; 
February  twenty  eight  alone,  and  all  the  reft  have  thirty  one. 

When  you  ran  divide  the  year  of  our  l^ord  by  4,  without  any  remainder,  it 
I*  tUcn  BiiTcxtilcor  Leap  Year,  ia  which  February  has  29  days. 


TABLES  IN"  COMPOUND  ADDITION.  47 

Seconds,  60  =  1  Minute. 

3600  =        60  =       1  Hour. 
86400  =     1440  =    24  ==     1  Day. 
604800  =  10080  =  168  =    7  =  1  Week. 
2419200  =  40320  =  672  =  28  =  4  =  1  Month. 
Sec.  Min.  h-         d.    h.      w.  d.h. 

31557600  =  525960  =  8766  =  365  6  =  52  1  6  =  1  Julian  year.* 

m.  sec. 
.11558154  =  525969  =  8766  —  365  6       9   14  =  1  Period,  year.t 
31556937  =  525948  =  8765  =  365  5    48  57  =  Tropical  year. J 

9.  Motion. 
60  Seconds     -         -         -         make  1  Prime  minute,  marked  "  ' 
60  Minutes         ,         -         -         -         Degree,  -  ** 

30  Degrees     -         -         -         -  Sign,      -         -         -         s. 

c^n^  J  \  I'he  whole' great  circle 

12  Signs,  or  360  degrees     -         -         ^      of  the  Zod,ack.§ 

Seconds,  60  =  1  Minute. 

3600  =         60  =       1  Degree. 
108000  =     1800  =    30  =     1  Sign. 
1296000  =  21600  =  360  =  12  =  Zodiack. 

10.  Land  or  Square  Measure. 
144  Inches  -  -  make  1  Square  toot. 

9  Feet         -  -  -  Yard. 

301  Yards,  or  >  p  , 

2721  Feet  J  " 

40  Poles         .  -  -  . .  Rood. 

4  Roods,  or  160  Rods,  >  ^         ^^^.^ 

or  4840  yards  J 

640  Acres        -  -  -  Mile. 

Inches,  144=  1  Foot. 

1296=  9=  1  Yard. 

39204=         2721=         30i=  1  Pole. 

1568160=       10890=        1210=         40=       1  Rood. 
6272640=       43560=       4840=        160=       4=      1  Acre.  ' 
4014489600=27878400=3097600=102400=2560=640=1  Mile. 

•  The  civil  folar  year  of  365  days,  being  (hort  of  the  true  by  5h.  48m.  48fec. 
occafioned  the  beginning  of  the  year  to  run  forward  through  the  fcafons 
nearly  i  day  in  4  years.  On  this  acrouct,  Julius  Ctfar  ordained  that  one  day 
fhould  be  added  to  February,  every  fourth  year,  by  caufing  the  a4th  day  to  be 
reckoned  twice;  and  hccaufe  this  24lh  day  was  the  fixth,  (fextilis)  before  the 
kalenda  of  March,  there  were  in  this  year,  two  of  thefe  fcxtiles,  which  gave  the 
name  of  Biflextile  to  this  year,  which  being  thus  corretStcd,  was  from  thence 
called  the  Julian  year. 

t  A  juft  and  equal  meafure  of  the  year  is  called  the  periodical  year,  as  being 
the  time  of  the  earth's  period  about  the  fun  ;  in  departing  from  any  fixed  point 
in  the  heavens,  and  returning  to  the  fame  again. 

I  The  fcveral  points  of  the  ecliptick  having  a  retrograde,  or  backward  mo- 
tion, the  equinox  will,  as  it  were,  meet  the  fun  ;  by  which  mean  the  fun  will  ar- 
rive at  the  Equinox,  or  firft  point  of  Aries,  before  his  revolution  is  completed, 
and  this  fpace  of  time  is  called  the  tropical  year. 

§  The  Zodiack  is  a  great  circle  of  the  fphere,  containing  the  12  figns,  througK 
Tvfilch  the  fun  pafle?. 


Ton  or  Load. 


48  TABLES  IN  COMPOUND  ADDITION. 

n.    Solid  Measure.* 

I72«  Indies  -  -  make  1  Foot. 

27  Feet         -  -  .  -  Yard. 

40  Feet  of  round  Timber,  or  } 

60  feet  of  hewn  Timber,    ) 

128  Solid  Feet,  i.  e.  8  in  length,  4}  i-.     ,    r  «t     < 

in  breadth  and  4  in  height,       I  ^""'^  <>^  ^ood. 

12.  Wine  Measure.! 

2  Pints         -  make  1  Quart,         marked  pts.  qts. 

4  Quarts  -  -      Gallon,  -  gal. 

10  Gallons     -  -  Anchor  of  Brandy,         anc. 

18  Gallons         -  -         Runlet,         -  -         run. 

311  Gallons  -  Half  an  Hogshead,       ihhd. 

42  Gallons         -  '         Tierce,  -  tier. 

63  Gallons  -  Hogshead,         -  hhd. 

2  Hogsheads  -  Pipe  or  butt,  P.  or  B. 

2  Pipes  -  -         Tun,         -  -       Tun 

Cubick  Inches. 
28f  =         1  Pint. 
57|  =        2  =       1  Quart. 
231  =         8  =         4  =       1  Gallon. 
9702  =    336  =     168  =    42  =  1  Tierce. 
14553  ==    604  =    262  =    63  =  li=  1  Hogshead. 
t9404  ==    672  =    336  =    84  =  2  =  11-=  1  Puncheon. 
29106  =  1008  =     604  =  126  =  3  =  2  =  li=  1  Pipe. 
38212  =  2016  =  1008  =262  =  6=z4=3=2  =  l  Tun. 

15.  Ale  or  Beer  Measure.| 

2  Pints         -  -     make  1  Quart,             marked  ptS.  qts. 

4  Quarts  -             -            Gallon,          -             -          gal. 

8  Gallons  -  -  -  Firkin  of  Ale  in  Lond.  A.  fir. 
S^Gallons  *             ■^           Firkin  of  Ale  or  Beer. 

9  Gallons  -  -  -  Firkin  of  Beer  in  Lond.  B.  fir. 
2  Firkins  -  -  Kilderkin,  -  -  kil. 
2  Kilderkins  -  -  Barrel,  -  -  bar. 
H  Barrel,  or  64  Gallons  Hogshead  of  Beer,           hhd. 

2  Barrels  -  -  Puncheon,         -         -       puii. 

3  Barrels  or  2  Hogsheads  Butt,         -  -  butt. 

*  By  Solid  Mcafurc  are  meafurcd  all  things  that  have  length,  breadth  and 
depth. 

f  All  Brandies, Spirits,  Perry,  Cider,  Mead,  Vinegar,Honey  and  Oil, arc  meaf- 
urcd by  Wine  Meafurc :     Honey  is  commonly  fold  by  the  pound  Avoirdupois. 

\  Milk  is  fold  by  the  Beer  quart. 

A  barrel  of  Mickarel,  and  other  barrelled  fifli,  by  hw  in  Maflachufctts,  is 
to  contain  not  lefs  than  30  gallons  ;  in  Connedlicut  and  New  York  the  Shad 
and  Salmon  Barrel  muft  contain  aoo  lb. 

In  England,  a  barrel  of  Salmon  or  Eels  is  42  gallons,  and  a  barrel  of  Herrings 
3a  gallons.  The  gallon> appointed  to  be  ufed  for  meafuring  all  kinds  of  Liquors, 
n  Ireland^'is  two  hundred  «ind  feventcen  cubick  indies,  and  fix  tenths. 


TABLES  IN  COMPOUND  ADDITION.  49 

Beer. 

Cubick  Inches.  I 
331  =       1  Pint. 
701  =      2  =       1  Quart. 

282  =       8  =      4  =  1  Gallon. 

2538  =    72  =    36  =  9=1  Firkin. 

5076  ^144=    72=  18=    2  =  1  Kilderkin. 

10152  =  288  =  144  =  36  =    4  =  2  =  1  Barrel. 

15228  =  432  =  216  =  54  =    6=3  =  li=  1  Hogshead. 

20304  =  576  =  288  =  72  =     8  =  4  =  2=  1^=  1  Puncheon, 

30456  =  864  =  432  =  108  =  12  =  6  =  3  =»=  2  =  li=  1  Butt. 

Ale. 
Cubick  Inches. 

351  =       1  Pint. 
70X  =      2  =       1  Quart. 
282  =      8  =      4=1  Gallon. 
2256  =     64  =    32  =     8  =  1  Firkin. 
4512  =  128  =    64  =  16  =  2  =  1  Kilderkin. 
9024  =  256  =  128  =  32  =  4  =  2  =  1  Barrel. 
13536  =  384  =  192  =  48  2^  6  =  3  =  li=  1  Hogshead. 

16.  Dry  Measure.* 

2  Pints          -  -          mpke  1  Quart,  marked  pfs.  qts. 

2  Quarts         -  -             -     Pottle,         -         -     pot. 

2  Pottles  -             -             Gallon     -         -         gal. 

2  Gallons      -  -             -       Peck,          -         -      pk. 

4  Pecks  -             -               Bqshel,    -         -         bu. 

2  Bushels      -  -             -       Strike,         -         -     str. 

2  Strikes  -             -              Coom,     -         -          co. 

2  Cooms       -  -             -        Quarter,      -         -      qr. 

4  Quarters  -             -                Chaldron,         -          ch. 
4i  Quarters  -             -          Chaldron  in  London. 

5  Quarters  -             -                Wey,        -         -     wey. 
2  Weys         >  -             -        Last,     -         -          last. 

Cubick  Inches. 

268|  =   1  Gallon. 
537f  =   2=1  Peck. 
2150|  =   8  =   4=1  Bushel. 
4300|  =  16  =   8=2=1  Strike. 
8601f  =32  =16  =4=2=1  Coom. 
172031  =  64=  32=  8=4=  2=  1  Quarter. 
86016  =  320  =  160  =  40  =  20  =  10  =  6  =  1  Wey. 
172032  =  640  =  320  =  80  =  40  =  20  =  10  =  2  =  1  Last. 


*  This  measure  is  applied  to  all  dry  goods,  as  Corn,  Seed,  Fruit,  Roots,  Salt, 
Sind,  Oysters  and  Coals. 

A  Winchester  bushel,  is  iS^  inches  diameter,  and  8  inches  dcrp, 

a 


m  COMPOUND  ADDITION 


COMPOUND  ADDITION 

IS  the  adding  of  several  numbers  together,  having  different  de- 
nominations as  Pounds,  Shillings,  Pence,  &c.  Tons,  Hundreds, 
Quarters,  &c. 

Rule.* 

I.  Place  the  numbers  so  that  those  of  the  same  denomination 
may  stand  directly  under  each  other. 

il.  Add  the  first  cohimn  or  denominalion  together  as  in  whole 
numbers  ;  then  divide  the  sum  by  as  many  of  the  same  denomina- 
tion as  make  one  of  the  next  greater,  setting  down  the  remainder 
under  the  column  added,  and  carry  the  quotient  to  the  next  supe- 
riour  denomination,  continuing  the  same  to  the  last,  which  add  as 
in  simple  addition. 

Examples. 

J,  Federal  Moke  v. 

1.  2,  3. 

E.  D.    d.    c.  m.  D.  c.  m.  D.  c.    m. 

7     3     8     9  6  49  18  7  375 

2  12  6  25  32  1  29  18 
9006  93  76  '7  12     5 

3  6     2  5  13  25  199  18     7 
7     14     0  8  97  2  30  01 


24      1      10     3 


!.  English  Money 


•  1. 

2. 

3, 

4. 

£  s. 

d. 

£  s. 

d.  qr. 

£ 

J.  d.     qr. 

£      s.        d.   qr. 

9  16 

10 

47  17 

6  2 

847 

11  11  2 

915  10  10  2 

7  10 

9 

-   3  9 

10  3 

491 

19  6  1 

64  8  9  1 

0  18 

6 

75  13 

9  1 

59 

6  10  0 

5  16  11  3 

5  11 

11 

4  11 

11  0 

747 

16   1  2 

419  2  10  2 

6  0 

8 

0  16 

8  2 

849 

12  11  3 

491  19  11  3 

6  9 

10 

17  6 

2  1 

741 

17  8  2 

762  17  6  1 

35     8 


As  the  denominations  of  Federal  Money  increase  iike  whoh. 
numbers,  in  a  ten  fold  ratio,  the  operation  is  the  same  as  in  whole 

*  The  reason  of  this  rule  is  evident  from  what  has  been  said  in  Simple  Ad- 
dition :  For, in  addition  of  money,  as  i,  in  the  pence  is  equal  to  4  in  the  far- 
things; I,  ill  the  shiUings,  to  la  in  the  pence;  and  i,  in  the  pounds,  to  ap  in 
the  shillings;  therefore,  carrying  as  directed,  is  the  arranging  the  money,  aris- 
ing from  each  column,  properly,  in  the  fcale  of  denominations  ;  and  this  rea'» 
soning  will  hold  good  in  the  addition  of  compound  numbers,  of  any  denomi- 
nation whatever. 


COMPOUND  ADDITION.  ^1 

numbers.  But  in  denominations  which  do  not  increase  in  the  same 
manner,  the  operations  are  somewhat  'tifferent.  ihus,  in  Ex.  1. 
of  English  Money,  I  find  the  sum  of  the  pence  to  be  64.  Now  51 
pence  are  4  shillings  and  6  pence  ;  therefore,  I  set  down  G  under 
the  pence,  and  carry  4  to  the  shillings,  which  I  then  find  to  be  68. 
But  68  shillings  are  3  pounds  and  8  shillings.  I  set  doxvn  the  8 
under  the  shillings,  and  carry  3  to  the  pounds,  and  the  sura  of  the 
pounds  is  35,  which  I  set  down.  The  su.n  of  the  whole  is  then  35 
pounds,  8  shillings  and  6  pence.  The  process  is  similar  in  each 
Example.  In  all  sums  oC  different  denominations,  the  student 
should  be  careful  to  find  the  numbers  by  which  the  denominations 
in  the  Table  increase,  for  by  Ihem  he  is  to  carry  from  one  denom- 
ination to  another. 


3.  Troy  Weight. 

I. 

2. 

3. 

li. 

oz.  fiut.  gr. 

ii.     oz.  fivt.  gr. 

th. 

oz.  pv>U  gr. 

767 

10  17  22 

649  11  19  20 

859 

9  15  20 

39 

6  9  17 

32  9  6  5 

437 

10  17  22 

417 

11  16  18 

841  10  11  19 

640 

11  6  0 

935 

9  17  19 

473  9  17  23 

738 

9  12  18 

478 

10  17  22 

764  11  8  9 

49 

0  16  17 

387 

9  16  15 

165  6  10  19 

584 

10  0  9 

3027    11    16    17 


la  the  1st  Ex.  I  find  the  8um  of  the  grains  to  be  113.  Now  113 
grs.  are  4  pwts.  and  17  grs.  because  24  is  contained  in  113,  four 
times,  and  17  is  the  remainder.  Then  1  set  down  17  under  the 
grs.  and  carry  4  to  the  pwts.  and  their  sum  is  96.  Now  96  pwts. 
are  4  oz.  and  16  pwts.  for  20  pwts.  make  1  oz. ;  therefore  I  set  16 
under  the  pwts.  and  carry  4  to  the  ounces,  which  makes  their  sum 
69.  But  59  oz.  are  4  lbs.  and  11  oz.  because  12  oz.  make  a  lb. ; 
therefore  I  set  down  11  oz.  and  carry  4  to  the  lbs.  which  makes 
their  sum  3027.  The  answer,  then  is  3027  lbs.  11  oz,  16  pwts. 
and  17  grs. 


4.  Avoirdupois  Weight. 

1. 

2. 

3. 

4. 

IL  oz,    Jr. 

Cwt.  qrs.  lb. 

T.  Civt.  qrs.  lb. 

r.  Cwt.  grs.  lb. 

oz.    dr. 

19  13  12 

17  3  19 

59   13  2  17 

91  17  2  25 

13  15 

21  9  6 

18  1  27 

6  17  1  21 

19  9  0  17 

10  12 

4  15  15 

9  2  9 

45  11  3  25 

14  13  2  0 

9  11 

22  10  5 

14  3  16 

57  16  2  19 

47  11  3  19 

14  0 

18  13  12 

12  0  6 

75  17  3  17 

69  0  1  0 

0  12 

6  11  10 

15  2  0 

6  19  0  26 

77  19  3  27 

15  11 

94  10  12 

5iJ  COMPOUND  ADDITION. 

6.  Apothecaries'  Weight. 


1. 

2. 

3. 

4. 

3  Bgr, 

5  3  9 

^'•• 

ft  3  3  9 

gr. 

ft 

3  3  B^r. 

9  1  17 

10  7  2 

19 

12  11  6  1 

15 

5 

9  3  2  13 

3  2  19  ■ 

6  3  0 

12 

4  9  10 

12 

4 

8  6  0  19 

6  1  17 

7  6  1 

17 

91  10  7  2 

16 

9 

10  5  2  12 

4  0  6 

9  5  2 

12 

4  8  12 

19 

6 

6  6  1  17 

5  2  12 

6  1  0 

16 

6  0  0  1 

10 

8 

9  4  0  0 

8  1  10 

9  3  2 

18 

4  9  2  1 

6 

7 

1  0  1  17 

33  2   1 

6.  Cloth  Measure 

. 

1. 

2. 

3. 

4. 

5. 

Yd.  qr.  n. 

/?.jE.  jr.  «. 

jE.jF/.  yr.  n. 

E,Fr.  qr.  n. 

Yds.  qr.  n. 

76  2  3 

91  3  2 

75  2  1 

49  3  3 

914  2  3 

3  3  1 

49  4  3 

7  1  3 

19  5  2 

49  2  1 

42  3  3 

6  2  3 

84  0  2 

24  2  1 

561  3  0 

67  2  2 

84  4  1 

76  2  3 

67  4  3 

84  0  2 

16  3  3 

7  0  0 

48  2  2 

48  2  2 

649^3  1 

49  2  2 

61  2  1 

7. ; 

9  2  3 

6  3  3 

617  1  3 

Long  Measure. 

1. 

2. 

3. 

4. 

5. 

It,  in.  bar. 

Yd 

.//. »«.    Po/.  ft. 

/«.  Milfur.pol. 

Deg,  mi.  fur. 

fol.  ft.    in.    he. 

9  11  2 

7 

2   11   12  11 

10   9  7  36 

759  bQ 

6 

29  15  10  2 

6  9  1 

4 

1  6    ! 

3  10 

9   7  3  19 

317  39 

1 

36  11  6  1 

7  0  2 

6 

0  10    1 

8  12 

11   4  1  24 

497  63 

7 

24  9  8  1 

8  10  0 

7 

2  9 

7  15 

6   6  5  12 

562  17 

0 

11  13  11  0 

9  6  2 

8 

1  10 

4  14 

9   4  6  9 

64  48 

5 

17  9  4  2 

7  10  2 

9 

2  11 

5  11 

11   5  1  10 

764  52 

4 

19  15  11  1 

8.  Time. 
1.  2.  3. 


IV.d.  h.     m.     s. 

Mo.d. 

/j. 

m. 

r.  >^. 

d. 

3^. 

mo.  IV.  d.    h.     tn.     s. 

3  6   22  57  42 

5  24 

19 

43 

19  10 

17 

67 

1 1  3  6.  23  29  55 

1  5  19  31  28 

4  27 

21 

35 

7  9 

27 

4 

8  1  1  19  45  38 

2  3  17  9  15 

9  18 

0 

12 

4  8 

16 

29 

9  2  3  17  18  19 

3  0  9  17  58 

4  19 

23 

19 

1  11 

14 

46 

10  2  5  11  50  13 

1  1  16  19  10 

8  11 

12 

13 

17  6 

9 

W 

9  2  1  16  18  17 

2  2  20  53  48 

9  19 

8 

29 

12  5 

20 

46 

9  3  6  18  17  59 

D. 

Motion. 

1. 

2. 

3. 

17*»  55'  48" 

25° 

49'  51" 

9s  2&°  35'  53" 

1  37  51 

6 

21  36 

10   0  18  31 

29  19  45 

19 

47  18 

4  17  13  42 

19  19  37 

25 

25  39 

6  19  60  0 

COMPOUND  ADDITION.  63 

10.  Land  or  Square  Measure. 


1. 

2. 

3. 

FoLfeet. 

in. 

Yds.  ft.    in. 

Acres,  rood.  poU  feet,     in. 

36   179 

137 

28  7   119 

756  3  37  245  120 

19  248 

119 

9  3     75 

29   1   28     93     25 

12     96 

75 

29  6   120 

516  3  31    128   119 

18   110 

122 

4  8      12 

37   1    19  218     20 

9  269 

24 

9  1    119 

61  0     0     92   103 

25  221 

143 

8  3     43 

Measure. 

191    1  25  129   136 

11.  Solid  M 

1 

' . 

2. 

3. 

Ton.  feet. 

in. 

rjif.  /^/r/. 

in. 

Cord.     feet.          in. 

29     36 

1229 

75     22 

1412 

37      119      1016 

12      19 

64 

9     26 

195 

9     110        159 

18     11 

917 

3     19 

1091 

48      127      1017 

19       8 

1001 

28     15 

1110 

8     111        956 

6       0 

523 

49     24 

218 

21          9         27 

17     39     1119  18     17     1225  9       28      1091 


12.  Wine  Measure. 


1. 

2. 

3. 

Tier 
37 

.  gal.     qts.  pts, 
36       3       1 

nbd.    gal.    qts. 
51       58       1 

t>ts. 

1 

Ton. 

37 

bbd,  gal.  qts. 

2     37     2 

9 

17     2     1 

27     39     3 

0 

19 

1      59      1 

35 

28     9     0 

9     18     0 

1 

28 

2       0     0 

32 

19     1     1 

0       9     2 

1 

19 

0     47     1 

9 

0     3     1 

16     24     1 

1 

37 

1      17     3 

12 

40     1      1 

1 
15. 

5       0     3 

0 

URE. 

14 

2     48     2 

Ale  and  Beer 

Meas 

1. 

2. 

3. 

A.B.    f.  gal, 

49     3     7 

£.  B.  fr.  gal. 

29     1     8 

Hhd. 
379 

gal. 

53 

qtu 

3 

26     2 

3 

19     3     5 

19 

0 

1 

9     0 

4 

16     0     3 

121 

37 

2 

17     3 

0 

9     1     8 

467 

19 

1 

27     1 

6 

14     2     0 

561 

16 

0 

19     3 

7 

17      1      5 

75 

0 

2 

16.   Dry  Measure. 
2. 


^rs. 

bu. 

P- 

qts. 

Bus. 

/• 

qts. 

pts. 

Cb. 

bu. 

/>• 

qts. 

64 

7 

3 

7 

37 

2 

5 

1 

37 

27 

3 

7 

9 

4 

1 

5 

19 

3 

7 

1 

6 

29 

1 

5 

19 

6 

2 

1 

\% 

2 

0 

0 

15 

30 

0 

0 

4 

0 

2 

0 

5 

1 

6 

1 

4 

11 

3 

9 

17 

3 

0 

6 

9 

0 

3 

0 

5 

0 

1 

0 

9 

5 

3 

4 

19 

3 

0 

1 

2 

0 

2 

1 

54 


COMFOUND  SUBTRACTION. 


COMPOUND  SUBTRACTION 

TEACHES  to  find  the  difference,  inequality,  or  excess,  between 
any  two  sums  of  divers  denominations. 

Place  those  numbers  under  each  other,  which  are  of  the  same 
denomination,  the  less  being  below  the  greater  ;  begin  with  the 
least  denomination,  and,  if  it  exceed  the  figure  over  it,  add  as  ma- 
ny units  as  make  one  of  tlie  next  greater  ;  subtract  it  therefrom  ? 
and  to  the  difference  add  the  upper  figure,  remembering,  always, 
to  add  one  to  the  next  superior  denomination,  for  that  which  you 
edded  before. 

Examples. 
Federal  Money. 


1. 


f 


$      c.   m. 
From  39     15     5 
Take  23     17     2 
Diff.      10     98     3 

c. 

18 

E. 

21 
10 

$ 
8 

7 

c.  m.              $ 
1     2              100 
5                       48 

c.  m. 

87     5 

Borrowed   100 
Paid               29 

m. 

Lent  200 
Received  145 

c. 

60 

He  mains  to  pay 

Due  tome 

$ 
Borrowed  3000 

c. 

$       c.     m. 
Lent  7159     12     8 

Paid        C    195 

at  \lllo     49 

several   j    247     37     5 


Received  C   245     37     5 
at  >3112     15 

several      i  2000 


limes. 


995     12     5      times. 


1092     92     0 


Paid  in  all  2552     99     0  Received  in  all 


Remains  to  pay  447     01     0     Remains  due 
English  Money. 


1. 

JC       s.  d.  qr. 

Borrowed  349     15  tj  1 

Paid     195     11  8  1 


2. 
£        s.    d.  qr. 
Lent  791        9     8     1 
Received   197     16     4     2 


iiem.  to  pay   154       3  10     0 
Proof 


Due  to  me 


"*  The  reafon  of  tbis  Rule  will  readily  appear,  from  what  was  faid  in  Simple 
.^abtraaion;  for  the  adding  depends  upon  the  fime  principle,  and  is  only  dif- 
•ferent.a?  the  numbers  to  b'^  fubtra'^ed  are  of  different  denominations. 


COMPOUND  SUBTRACTION.  5^ 

In  the  1st  Ex.  of  English  Money,  I  take  1  qr.  from  1  qr.  and  set 
down  0,  the  remainder.  Because  I  cannot  take  eight  from  6  pence. 
I  add  to  6,  12  pence  which  make  a  shilling,  and  from  18  take  8,  and 
set  down  10,  the  difference.  As  I  added  12  pence  =  1  shilling  to 
the  upper  pence,  I  now  carry  1  shilling  to  the  lower  shillings,  and 
take  12  from  15,  and  set  down  3,  the  remainder.  The  rest  of  the 
process  is  evident.  It  is  obvious  that  a  similar  course  must  be  pur- 
sued in  the  Examples  under  the  several  weights  and  measures. 

3.  Tnoy  Weight. 

I.  2.                                     3. 

lb.  oz.  pwt.    dr.  lb.    oz.  pwt.  gr.             lb.  oz.  pwt.    gr. 

Bought  749    5     13     16  189     8     12     10           543    3      9     13 

Sold          96    9     19     13  148     4     16     19           179     1     16     10 


Rem.  652    ' 

7     14 

[    03 

1. 

lb.  oz. 

Bought  7     9 

Sold        3   12 

dr. 
12 

9 

4.  Avoirdupois  Weight 
2.                      3. 

C.  rr.  lb.          T.cwt.qr.  lb. 
8  2   13          6   13   1    12 
4   1    15          1    12  2  17 

4. 
T.cwt.qr.  lb.  oz.    dr. 
9    11    3   17      5   12 
3   12   1    19   10     9 

Rem.  3   13 

3 

13 
16 

1. 

Ife     3    3     9 
71     9.3     1 
37     8     4     1 

5.  Apothecaries'  Weight. 

2.                                     3. 
ife      3     5    9     gr.          ft     3     3    9   gr. 
65     10     6     2     10           84     1      1      1        1 
31        8     4     2       9           65     9     3     1      17 

34     0     6     2 

17 

I 

Pol. 
21 

9 

1. 

Yd8.qr.    n. 
35      1      2 
19      1     3 

6.  Cloth  Measure. 

2.                                 3. 
:.  E.    qr.    11.                E.  Fl.     qr. 
167     3      1                 765      1 
>91     3     2                149     2 

4. 
n.               E.Fr.    qr.  n, 
3                549     4     2 
1                197     4     '^ 

15     3     3 

1. 

Yds.  ft.  in. 

28  2   10 
17  2   11 

7.  Long  Measure. 
2.                   3. 
ft.  in.       Mil.fur.poJ.       Beg. 
11   9          76  3   11          38 
13  8          27  3  21           19 

4. 
m.  fun  p.  yds,ft.in.bar. 
41  3  29  2   1    7  2 
35  5  31   3  1   9   1 

10  2   11 

1. 
Mo.  d.    h.    m. 
G    17    13  27 
1   21    16  41 

s. 

19 
'35 

44 

8.  Time. 
2.                  3. 
Mo.w.d.    h.       Y.mo.  d. 
9  2  5   15       7  3   13 
4  3  5   15       4  2   19 

4. 
Y. mo.w.d.    h.    m.      s, 
48  9  2  5   19  27  31 
19  9  3  4  20  19  49 

4  25  20  45 

5G  '      COMPOUND  SUBTRACTION. 


1. 
79°  21'  sr 

41   41  52 

9.  Motion. 

2. 
6s  11°  12'  48" 
3   8   39  29 

Mea 

A. 

56 
29 

J. 
4s  19°  41'  22'^ 
1   22   19  46 

10. 

1. 
A.  R.  Pol. 
29   1   10 
24  1  25 

Land  or  Square 

2. 
A.  R.  Pol. 
29  2   17 
17  1  56 

SURE. 

3. 
R.  Pol.   ft.    i"n. 
3  19   27  110 
0  21  210  129 

I. 
Tons.  ft.   in. 
49  19  1100 
38  36  1296 

11.  Solid  Measure. 

2. 
Ydu.   ft.    in. 
79   11   917 
17  25  1095 

3. 
Cords.   ft.    in. 
349   97  1250 
192  127   1349 

12.  Wine  Measure. 

1.  2.  3.  4. 

Hhd.  gaK  qts.  pts.        Tier.   gal.  qts.        Hhd.     gal.  qts.  Tun.  hhd.  gal. 

79  21      2      1  19      17      1  375     41     2  532      1      19 

38  61      3      1  12     29     2  197     36     3  197      1     47 


13,  Ale  and  Beer  Measure. 

1.  2.                                    3. 

A.B.  fir.  gal.  qts.  B.B.  fir.  gal.  qts.  pts.             Hhds.   gal.  qts. 

39      1      2      1  21      3     5     2     0              769      17      1 

24     3     6     2  19      1      7     2      1              391     42     3 


14.  Dry  Measure. 


1.  2.  3. 

Qa.    bu.  pk.  qts.  Bu.  pk.  qts.  pts.  Ciial.  bo.   pk.  qts. 

56     2     2      1  91      1     3     2  39  12     2     1 

39     3     1     2  29     2     1      1  24  25     3     2 


PROBLEMS.  #  «T 

PROBLEMS 

nESULTING    FROSl    A    COMPARISON    OF    THE    PRECEDING    RULES. 

Prob.  1.  Having  the  sum  of  two  numbers,  and  one  of  them  giv- 
eti,  to  find  the  other. 

Rule.  Subtract  the  given  number  from  the  gi  vert  sum,  and  the  re- 
mainder will  be  the  number  required. 

Let-SSS  be  the  sum  of  two  num-      From  288  the  Sum, 
hers  ;  one  of  which  is  1 15,  the  oth-     Take   1 15  the  given  number. 
<'r  is  required  ?  Rem.   1 73  the  other. 

pROB.  2.  Having  the  greater  of  two  numbers,  and  the  differ- 
ence between  that  iind  the  less  given,  to  find  the  less. 

Rule.     Subtract  the  one  from  the  other. 

Let  the  greater  number  be  325,  and     From  325  the  greater, 
the  difference    between  that  and  the     Take  198  the  difference. 
Other,  198:     What  is  the  other?  Rem.  127  the  less. 

Prob.  3.  Having  the  least  of  two  numbers  given,  and  the  dif- 
ference between  that  and  a  greater,  to  find  the  greater. 

Rule.     Add  them  together. 

p.         ^    127  the  less  number. 
uiven  ^    193  ^i^^  difference. 

Sum  325  the  greater  number  required. 

Prob.  4.  Having  the  sum  and  difference  of  two  numbers  given, 
to  find  those  numbers. 

Rule.  To  half  the  sum  add  half  the  difference,  and  the  sum  U 
the  greater,  and  from  half  the  sum  take  half  the  difference,  and  the 
remainder  is  th«  less.  Or,  from  the  sum  take  the  difference,  and 
half  the  remainder  is  the  least :  to  the  least  add  the  given  differ- 
ence, and  the  sum  is  the  greatest. 

What  are  those  two  numbers,  whose  sum  is  48,  and  difference  14  ? 
2)48  2)14  24-f  7=31  the  jrreater,and  24— 7=17 the  less. 
i.s„m=24  }m.^  Ov46'—l4-^2=\l,  &  17+14=3J. 

This  rule  is  obvious  on  considering  any  exanaple  in  the  following 
manner.  Thus,  let  the  two  numbers  be  S2  an<l  46  ;  their  sum  is 
46-f-32,  and  their  ditference  is  4<;— 32.  Now  46-f  32-~2-|-46— 3:J 
>^2=4G-f 324^46— 32--r-e=46"-f-4G--f-2=J6,    the    gfreater.       And 


4G-|-32-~2— 46-{-32~-2  =4t)-f32~46-f  32-^-2  =  32-r32-4-2=  32, 
the  less.  For,  in  the  first  case,  the  32  to  be  subtracted  from  32, 
leaves  nothing  ;  and,  in  the  latter, .the  46  is  balanced  by  the  other 
46,  and  you  have  only  32-j-32-r-2=32. 

The   preceding  and   following  probjoms  are   evident   from  the 
rules  of  Additiuri  and  Subtraction,  Multiplication  and  Division. 

Prob.  5.     Having  the  product  of  two  numbers,  and  one  of  theffs 
given,  to  find  the  other. 

Rule.     Divide  the  product  by  the  given  numb^^.r,  and  the  quotient 
wfll  be  the  number  required. 

If 


5S  P  MiOliLEMS. 

Let  the  product  of  two  numbers  be  288  8)28b' 

and  one  of  tliem  8  ;  I  demand  the  other  ?         Answer^         36 

PiipB.  G.     Having  the  dividend  and  quotient,  to  find  the  divisor 

Rule.     Divide  the  dividend  by  the  quotient. 

Cor.  Hence  we  s^et  another  method  of  proving'  Division. 
C'von    5^^^  ^^^®  Dividend.  30)288(8  Divisor. 

}    36  the  Quotient.                                   288 
Required  the  Divisor. 

pROB.   7.     Having  the  Divisor  and  Quotient  given,  to  find  the 

Dividend. 

Rule.     Multiply  them  together. 

^.  ^8  the  Divisor.  JG 

uiven    <  oo  .1     /^     .•     ,  o 

^  3G  the  Quotient.  8 

Required  the  Dividend.  -»--^ — 

288  the  dividend. 

By  a  due  conrsideralion  and  apphcation  of  these  Problems  only 

many  questions  (of  which  kind  are  some  of  the  following)  may  be 

resolved  in  a  short  and  elegant  manner,  although  some  of  them  are 

generally  supposed  to  belong  to  higher  rules. 

APPLICATION    OF    THE    PRECEDING    RULES. 

1.  The  least  of  two  numbers  is  19418,  and  the  difierence  be= 
tween  them  is  2384 :   What  is  the  greater,  and  sum  of  both  ? 
19418-f2384=21802  greater,  and  1941.8-f 21802=41220  sum. 

2  Suppose  a  man  born  in  the  year  1743  ;  when  will  he  be  77 
years  of  age  ?  1743-f77=1820  Ansmer. 

3.  Wbat  number  is  that,  which,  being  added  to  19418,  will 
make  21802?  2384  c^n5. 

4.  Gen.  Washington  was  born  in  1732  ;  what  was  his  age  in 
1799  ?  67  Ms, 

6.  America  was  discovered  by  Columbus  in  1492  and  its  inde- 
pendence declared  in  1776  :  How  many  years  elapsed  betweeti 
those  two  eras  ?  284  A7is. 

6.  1  he  Massacre  at  Boston,  by  the  British  troops,  happened 
March  5th,  1770,  and  the  Battle  at  Lexington,  April  19lh,  1775; 
How  long  between  ? 

April  19th,  1775~March  5th,  177C=5  y.  1  m.  14  d.  Jlns. 

7.  Gon.  Burgoyne  and  his  army  were  captured  October  17th, 
1777,  and  Karl  Cornwallis  and  his  army,  October  19th,  1781  : 
What  space  of  time  between  ?  4  years  and  2  days.  Ans. 

8.  The  war  between  America  and  Kngland  commenced  April 
I9th,  1775,  and  a  general  peace  took  place  .Tannary  20th,  1783: 
How  lonix  did  the  war  continue  ?  "7  y.  9  m.  1  d.  Ann. 

9.  A,  h,  C  and  D  ptircliased  a  quantity  of  goods  in  partncrsbip; 
A  paid  £  12  10.9.  a  doilHr*  and  a  crnvvnt  piece  ;  P),35s.  C,  29s.  lOi/. 
and  D,  79*;.  ;    What  did  the  good:,  co.sl  ?  .i/n.v.  i.'  \0  hi  1. 


l^KOBLEMS.  59 

10.  A  man  borrowed,  at  different  limes,  these  several  sums,  viz. 
£29  5s.  £18  17s.  Gd.  £45  I2s.  £98,  3  dollars,  one  crown  piece 
and  an  half;  Pray  how  much  was  he  in  debt?      Ans.  £  193  2  G 

11.  There  are  four  numbers  ;  the  first  317,  the  second  912,  the 
third  1229,  and  the  fourth  as  much  as  the  other  three,  abating  97  : 
What  is  the  sum  of  them  all  ?  Ans.  4819. 

12.  Bought  a  quantity  of  goods  for  -£125  10s.  paid  for  truckage 
45s.  for  freight  79s.  6d.  for  duties  35s.  lOd.  and  mv  expences  were 
j3s.  9d. :   What  did  the  goods  sfan<l  me  in  ?  Ans.  £  136  4  1. 

13.  A  Gentleman  left  his  son  £  1725  more  than  his  daughter, 
vhose  fortune  was  15  thousand,  15  hundred  an/l  15  fiounds:    What 

was  the  son's  portion,  and  what  did  the  whole  estate  amount  to? 
Ans.   The  son's  fortune,  £  J8240,  and  the  whole  estate  £34755. 

14.  A  mercfiant  had  6  debtors,  who  together  owed  him  £2917 
10s,  Gd.  A,  B,  C,  D  and  E,  owed  him  £  1675  13s.  9t/.  of  it:  What 
was  F's  debt  ?  Ans.  £  1241  16  9. 

^  15.   What  is  the  difference    between   £1309  7s.    Id.   and   the 
amount  of  £  345  13s.  4d.  and  £  571  4s.  Sd.  1         Ans.  £392  9  1 

16.  A  merchant,  at  his  first  engaging  in  trade,  ovyed  £937  15s. 
he  had  in  cash  £1755  3s.  Gd.  in  goods  £459  12s.  ?hI.  in  good  debts 
£197  16s.  and  he  cleared  the  first  year  £219  19s.  \Qd.  *"  What  was 
the  neat  balance  at  the  year's  end  ?  Ans.  £1724  16  7. 

17.  What  sum  of  money  must  be  divided  between  12  men,  so  as 
that  each  may  receive  £155  ?  £  1860  Aus. 

18.  What  number  must  I  multiply  by  9,  that  the  product  may 
b£  675  ?  75  Ans. 

19.  A  privateer  of  175  men  took  a  prize,  which  amounted  to 
i^59  per  man,  beside  the  owner's  half:  What  was  the  value  of  the 
prize  ?  £20650  Ans. 

20.  What  is  the  difference  betwepn  thrice  five,  and  thirty,  and 
thrice  thirty  five  ?  6Q  Ans. 

21.  The  sum  of  two  numbeis  is  750;  the  less  248:  What  i=! 
their  difference  and  product  ?  ditf.  254,    124496  product. 

22.  What  is  the  difference  between  six  dozen  dozen,  and  half  a 
dozen  dozen  ;  and  what  is  their  proviucl,  and  the  quotient  of  the 
greater  by  the  less  ? 

Ans.  792  difference,  62208  product,  and  12  quotient. 

23.  There  are  two  numbers  ;  the  greater  of  them  is  25  times 
78,  and  their  difference  is  9  times  15  ;  their  sum  and  product  are 
required. 

Ans.  1950  the  greater,  1815  the  less,  3765  the  sum,  and 
3.^39250  the  product. 

24.  A  merchant  began  trade  wilh  £25327;  for  six  years  togetii' 
or,  he  cleared  £1253  per  annum  ;  the  next  5  years,  he  cleared 
£1729  per  annum;  bijt,  the  last  4  years,  had  the  misfortune  fo 
[o,-:e  £3019  per  annum  :   What  was  he  worth  at  the  15  years'  end  ? 

Ans.  £29414. 

25.  If  a  man  r^pends  £  IC2  in  a  voar:  What  is  that  per  calendar 
:r.Gril}>  ^  ^  '  £16  An^. 


# 


60  REDUCTION- 

26.  If  the  Federal  Debt,  which  is  42  million  dollars,  be  equally 
divided  between  the  13  States  :   What  will  be  the  share  of  each  ? 

Ans.  3230769^3  dollars. 

27.  If  9000  men  march  in  a  column  of  750  deep :  How  many 
march  abreast?  ]2  Jlns. 

28.  What  number,  deducted  from  the  32d  part  of  3072,  will 
leave  the  96th  part  of  the  same  ?  64  Jlns. 

29.  What  number  is  that,  which>  multiplied  by  3589,  will  pro- 
duce 92050672  ?  25648  Ans. 

30.  Suppose  the  quotient  arising  from  the  division  of  two  num- 
bers to  be  5379,  the  divisor  37625:  What  is  the  dividend,  if  the 
remainder  came  out  9357  ?  202394232  Ans. 

31.  There  is  a  certain  number,  which  being  divided  by  7,  the 
quotient  resulting  multiplied  by  3,  that  product  divided  by  5,  irom 
ihe  quotient  20  being  subtracted,  and  30  added  to  the  remainder, 
the  half  sum  shall  make  35:  Can  you  tell  me  the  number? 

700  Ans. 

32.  A  sheepfold  was  robbed  three  nights  successively  ;  the  fir«t 
night,  half  tha  sheep  were  stolen,  and  half  a  sheep  m.ore  ;  the 
second  half  the  remainder  were  lost,  and  half  a  sheep  more  ;  the 
last  night  they  took  half  what  were  left  and  half  a  sheep  more  ; 
by  which  time  they  were  reduced  to  30:  How  many  were  there 
at  first  ? 

Begiti  with  30,  and,  reckoning  back  fiom  the  last  night  (o  Ih^ 
first,  you  will  find  that  31  were  stolen  the  3d  night,  62  the  2d,  atid 
124  the  first.  Ans.  247. 

33.  Two  boys,  A  and  B,  had  850  chesnuts  between  them;  but 
A  had  150  more  than  B  :  How  njany  had  each. 

85C-r-2=425  half  sum,  and  150-t-2^75  half  di(f.  ;   thtn  425-^75 
=500  A's,  and  425—75==350  B's. 

S4.  What  number  added  to  the  27th  part  of  6615,  will  make  570* 

325^  Ans. 


REDUCTION 

T]'..\('\l{]S  t>^  bring  nntnlirrs  of  one  (!onomuia'.i(in  to  ulhors  ol 
dillVrerit  dtnominatiotif?,  retaijiing"  the  same  value, 
h  is  of  /ci'o  feorts,  \\a   Descending  and  Ascending. 


RKDUCTiO::  DIISCENDIKG 

Toiclir-^  to  c!iarge  nnmiiors  from  i  hia^her  to  a  lower  dcnorai- 
nation.     It  is  performed  by  mnltiplic.a'.irtj. 


REDUCTION.  61 

Rule* 

Multiply  the  highest  denomination  given,  by  so  many  of  the 
!iext  less  as  make  one  of  that  greater,  and  thus  continue  until  yod 
have  brought  it  down  as  low  as  your  question  requires. 

Proof.  Change  the  order  of  the  question,  and  divide  your  last 
product  by  the  last  multiplier,  and  so  on. 

Note.  From  this  rule  and  Case  Vf.  of  Simple  Multiplication,  it 
appears,  that  Federal  Money  is  reduced  from  higher  to  lower  de-^ 
nominations  by  annexing  as  many  cyphers  as  there  are  places  from 
the  denomination  given,  to  that  required  ;  or,  if  the  given  sum  be 
of  different  denominations,  by  annexing  the  several  figures  of  all 
the  denominations  in  their  order,  and  continuing  with  cyphers,  (if 
necessary,)  to  the  denomination  required  ;  or,  what  amounts  to  the 
same  thing,  by  reading  the  whole  number  from  the  left  to  the  re- 
quired denomination,  as  one  number  in  the  required  denomination. 

Examples. 

1.  In  3  eagles  2  dollars,  how  many  mills  ?  Ans.  32000  m. 

2.  In  91  dollars  75  cents,  how  many  cents  ?  Am.  9175  c. 
.3.  In  50  eagles,  how  many  dollars  ?  Ans,  600  D 

4.  In  44  dollars,  1  cent,  4  mills,  how  many  mills  ? 

5.  In  9  dollars,  31  cents,  7  mills,  how  many  mills? 

6.  How  many  cents  in  39  dollars  5  cents  ? 

7.  In  28  dollars  17  cents,  5  mills,  how  many  mills? 

8.  In  £27  15s.  ^)(l.  2qrs.  how  many  farthings? 

£     s.     d.     qr. 
27     11      9     2 
multiplied  by  20=shillings  in  a  pound. 


55b=shininss 


'by     I2=pence  in  a  shilling. 

G6e9=pence. 
~^y       4=forthings  in  a  penny. 


Ans.     =2GG78  farthinsrs 


jYote.  In  multiplying  by  20,  I  added  In  t!ie  15s.  by  It,  the  $<3. 
And  by  4,  the  2ijrs. which  must  always  be  done  in  like  cases. 

To  prove  the  ab  )ve  question,  change  the  order  of  it,  and  it  wi^ 
stand  thus  :   In  20678  farthings  how  many  pounds  ? 

*  Tlie  reason  of  this  Rule  is  exceedingly  obvious ;  for  pounds  are  br'ouj'ht 
into  shillings  by  multiplying  tlicm  by  20  ;  shillings  into  pence  by  multiplying 
them  by  12;  and  pence  into  farthings  by  multiplying  them  by  4;  and  the 
contrary  by  division  ;  and  this  will  be  trnr  in  the  rfluction  of  numbers  con- 
■iisting  of  any  dcnomii.ation  whatever  Die  rule  for  Reduction  ascending  is 
simply  the  reverse  of  this,  and  equally  evident. 


6^1  K£DUCT|ON. 

4)26678 
12)6669  2qrs. 
2|0)55|5  9  d. 


Answer,  £27   15  9  2 

C.  In  £36   12s.  lOtl.   Iqr.  how  many  fartljingg  ?  Ans.  35177. 

10.  In  £95   Us.  5(1.  3qrs.  how  m^ny  farthings  ?  Ans.  91751. 

1 1.  Jn  £  719  9s.   1  Id.  how  many  halfpence  ?  Ans.  345358. 

12.  In  29  guineas,  at  28s.  how  many  pence  ?  Ans.  9744. 

13.  In  37  pistoles,  at  22s.  how  many  shillings,  pence,  and  farthings  ? 

Ans.  814s.  9768d.  39072qrs. 

14.  In  49  half  Johannes,  at  48s.  how  many  sixpences  ?     Ans.  4704-. 

15.  In  473  French  crowns,  at  6s.  8d.  how  many  threej>ences  ? 

An^.  126131 
\Q.  In  53  moidores,  at  36s.  how  many  shillings,  pence  and  farthings? 

Ans.  1908s.  22896d.  91584qrs. 

17.  In  ^29  how  many  groats,  threepences,  pence,  and  farthings  ? 

Ans.   1740  groats,  2320  threepences,  6960d.  27840qr3. 

18    Reduce  47  guineas  and  one  fourth  of  a  guinea  into  shillings^ 

sixpences,  groats,  threepences,  twopences,  pence  and  farthings. 

Ans.  1323  shillings,  2046  sixpences,  3969  groats,  5292  three- 
pences, 7938  twopences,  15876  pence,  and  63504  qrs. 

REDUCTION  ASCENDLYG 

Teaches  to  change  numbers  from  a  lower  to  a  higher  denorni 
nation.     It  is  performed  by  division. 

Rule. 

Divide  the  lowest  denomination  given,  by  so  many  of  that  name, 
as  make  one  of  the  next  higher,  and  thus  continue  till  you  have 
brought  it  into  that  deno.nination  which  your  question  requires. 

Note.  From  this  rule  and  the  note  under  Case  II.  of  Simple  Di- 
vision, it  appears,  that  Federal  Money  is  reduced  from  lower  to 
iiighor  deiiominations  by  cutting  off  as  many  places  as  the  given 
denomination  stands  to  the  right  of  that  required  ;  the  figures  cut 
off  belonging  to  their  rc^pective  denominations. 

Examples, 

1.  How  many  eagles  in  42000  mills  ?  Ans.  4  E.  jj  2 

2.  In  3175  cents,  how  many  dollars  ?  Ans.  $  31  75  c. 
.1.  In  500  dollars  how  many  Eagles  ?  Ans.  SQ 
4.  In  44^4  mills,  how  many  dimes  ? 

i».  In  9317  mills,  how  many  dollars  ? 

-;.  How  many  dollars  in  28175  mills? 

T.   In  547325  (\irthingSj  how  many  pence, 'shillings,  and  potjiuls' 


REDUCTION.  68 

Farthings  in  a  penny  =  4)547325 

Pence  in  a  shilling  =  12)136831      1  qr. 

Shillings  in  a  pound  =  2|0)1140|2     7d. 

£570  2s.  7d.  1  qr. 
Ans.   136831d.   11402s.  and    £570. 
Note.    The  remainder  is  always  of  the  sarae  name  as  the  dividend. 

8.  Bring  35177  farthings  into  pounds. 

9.  Bring  91751  farthings  into  pence,  &c. 

10.  Bring  345358  halfpence  into  pence,  shilling«<,  and  pounds. 

11.  Reduce  9744  pence  to  guineas,  at  28s.  per  guinea. 

12.  In  39072  farthings,  how  many  pistoles,  at  22s. 

13.  In  4704  sixpences,  how  many  half  Johannes  ? 

34.  In  12613i  threepences,  how  many  French  crowns,  at  6s.  Sd.-? 
15.  In  91584  tarlhings,  how  many  moidores,  at  36s.  ? 

16.  In  27840  farthings,  how  many  pence,  threepences,  groats, 
shillings  and  pounds  ? 

17.  In  63504  farthings,  how  many  pence, twopences,  threepenc- 
es, groats,  sixpences,  shillings  and  guineas  ? 

Note.  The  preceding  questions  may  serre  as  proofs  to  those  in 
Reduction  descending. 

REDUCTION  DESCENDING  AND  ASCENDING. 
1.  Money. 

1.  In  £97  how  many  pence  and  English  or  French  crowns,  at. 
Gs.  8d.  ?  Ans.  23280d.  and  291  crowns. 

2.  In  947  English  crowns,  at  6s.  8il.  how  many  shillings  and  Eng- 
lish guineas?  Ans.  6313s.  4d.  and  225  guineas  13s.  4(i. 

3.  In  519  English  half  crowns,  how  many  pence  and  pounds? 

Ans.  20760d.  and  £  86   10s. 

4.  In  1259  groats,  how  many  farthings,  pence,  shillings,  and  guin- 
eas ?  Ans.  20144qrs.  5036d.  419s.  8d.  and  14  guin,  27s.  8d. 

5.  In  75  pistoles,  how  many  pounds  ?  Ans.  £  82   10s. 

6.  In  735  French  crowns,  how  many  shillings  and  French  guin- 
eas, at  26s.  8d.  ?  Ans.  4900s.  and  183  guin.  24s. 

7.  In  5793  pence,  how  many  farUiings,  pounds,  and  pistoles? 

Ans.  23172qrs.  £24  2s.   9(1,  and  21  pistoles,  20s.  9d. 

8.  In  £  99,  how  many  shillings,  and  half  Johannes,  at  48s.  ? 

Ans.  1980s,  and  41  half  joes.  12s. 

9.  In  £  179,  how  many  guineas  ?  Ans.   127  guin.  24s. 

10.  In  £  345  how  many  moidores  ?  Ans.  191  moid.  24s. 

11.  In  59  half  joes,  37  moidores,  45  guineas,  63  pistoles,  24  Eng- 
lish crowns,  and  19  dollars  ;  how  many  pounds,  half  joes,  moidores, 
guineas,  pistoles,  English  crowns,  dollars,  shillings,  pence,  and  far 
things  ? 

Ans.  £354  4s.  147  half  Joes,  28s.  196  moidores,  28s.  253  guin- 
eas, 322  pistoles,  1062  English  crowns.  4-,  1 180  dollar?,  4^  ':p>p--% 
shillings,  85008d.  and  340032qrs. 


64 


REDUCTION. 


When  it  is  required  lo  know  how  many  sorts  of  coin,  of  differ- 
ent values,  and  of  equal  number,  are  contained  in  any  number  of 
another  kind  ;  reduce  the  several  sorts  of  coin  into  the  lowest  de- 
nomination mentioned,  and  add  them  together  for  a  divisor ;  then 
reduce  the  money  given,  into  the  same  denomination,  for  a  divi- 
dend, and  the  quotient,  arising  from  the  division,  will  be  the  num- 
ber required. 

Note.     Observe  the  same  direction  in  weights  and  measures. 

1.   In  275  half  Johannes,  how  many  moidores,  guineas,  pistoles, 
dollars,  shillings  and  sixpences,  of  each  the  like  number  ? 
A  moidore  is  36s.       }  „r>    •  275  half  joes. 

that  is  I  72  sixpences.  48  sl.il.  in  ajohau. 


A  guinea  is  28s. 
that  is 


A  pistole  is  22s. 
that  is 

A  dollar  is  6s. 

that  is 

One  shilling  has 


>  56  ditto. 


44  ditto. 


2200 
1100 

13200  shillings. 


sixp.  in  a  shill. 
dividend=26400  sixpences. 


^  12  ditto 

2  do.  187)26400(141  of  each  and  33  ?ixp.  or 
1  do.  " 


Divisor=187  sixpences. 
2.  A  Gentleman  distributed  £37  10s.  between  4  poor  persons, 
JQ  the  following  manner,  viz.  that  as  often  as  the  first  had  209.  the 
second  should  have  15s.  the  third,  10s.  and  the  fourth,  5s.     What 
did  each  person  receive?  Ans.  The  first  man  J£15,  second 

£11  5s.  third  £7  10s.  fourth  £3  15s. 


2.   Troy  Weight. 
1.  How  many  grs.  in  a  silver  bowl,  that  weighs  31b.  10  oz.  12 


pwt. 


ft  oz.  pwt. 
3   10   12 
12  ounces  in  a  pound. 


46  ounces. 

20  pennyweights  in  an  ounce. 

932  pennyweights. 
24  grains  in  one  pwt. 

3728 
1864 


i'roof.   24)22368  grains,  answer. 


2|0)93|2 
12)46—12  pwt. 
Jfe  3—10  oz. 


REDUCTION.  65 

t.  In  4870ZS.  how  many  pwts.  and  grs.  ? 

Ans.  9740pwt.  and  233760gr. 

3.  In   13  ingots  of  gold,  each  weighing  9oz.  Spwt  how  many 
grains  ?  Ans.  67720gr. 

4.  In  97397grs.  how  many  pounds  ?  Ans.  16ife  lOoz.  I8pwt   5gr. 

5.  How  many  rings,  each  weighing  5pwt.  7gr.  may  be  made  of 
lib.  5oz.  16pwt.  2gr.  of  gold.  Ans.  158. 

3.  Avoirdupois  Weight. 

Cwt    qrs.   ib      oz. 
1.  In     91      3     17     14  how  many  ounces? 
4 

367  quarters.  Proof. 

28  16)164702 


2943  28)10293  14oz. 
735  - 
4)367  171b. 


10293  pounds.  

16  Cwt.  91  3qrs. 


61762 
10294 


164702  ounces!. 

2.  In  12  tons,  15cvvt.  Iqr.  19!b.  6oz.  12dr.  how  many  drams  ? 

Ans.  7323500dr. 

3.  In  241b.  1  loz.  9dr.  how  many  drams  ?  Ans.  6329dr. 

4.  In  44800  pounds,  how  many  drams  and  tons  ? 

Ans.  1 1468800dr.  and  20  tons. 

5.  In  281b.  Avoirdupois  how  many  pounds  Troy  ? 

28 
7000  grains  in  1  lb.  Avoirdupois. 


f  ;'^- '"J  =576!0)19600!0(34}fe 
Ub.lr.^  i  ^j,728  6.  In 


47lb.  9oz.  13pwl.  17gr.  Troy, 
how  many  pounds  Avoirdupois  ? 


2.320  '47     9      13     1 

2304  12 

160  673 

12  20 

,76|0)192|0(Ooz.  11473 

20  24 


.^76|0)3840|0)6pwt.  45899 

3456  22947 


3840  carried  over.     275369  carried  ovfi-n-. 
I 


0(5 


REDU 

CTION. 

Brought  over.     3840 
24 

7l000)275l369(39fe  Br'ghl  OTCr. 
21 

1536 

66 

768 

63 

576|0)9216|0(16gr. 
576 

2369 
16 

3456 

14214 

3456 

2369 

7  [000)37 1 904  (5oz. 
35 

2904 

16 

17424 

2904 

7|000)46|464(64ff|(lr„ 
42 

4464 

4.  Apothecaries'  Weight. 
1.  How  many  grains  are  there  in  37ife  63  ? 

ft     5  Proof. 

37     6  2|0)21600|G 

12  

3)10800 

450  ounces.  , — 

8  8)3600 


3600  drams.  12)450 

3  

37  ife  65 

10800  scruples*  

20 


Ans.  216000  grains. 

2.  In  9ft  83  13  29  19gr.  how  many  grains  ?     Ans.  55799gi'. 

3.  In  55799  grains,  how  many  pounds,  &c.  ? 

Ans.  91fe  83  13  29  19gr. 

5.  Cloth  Measure. 
1.  In  127  yards,  how  many  quarters  and  nails  ? 

4  Proof. 

Ans.    508  qrs.  4)2032 

4  4)508 

Ans.  2032  naili.  127  yards.. 


REDUCTION- 


er 


5.  In  9173  nails,  how  many  yards  ?         Ans.  573yds.  Iqr.  In, 

3.  In  75  ells  English,  how  many  quarters  and  nails  ? 

Ans.  2^75qrS.  1500n, 

4.  In  56  ells  Flemish,  how  many  quarters  and  nails  ? 

Ans.  168qrs.  672n. 

6.  In  39  ells  French,  how  many  quarters  and  nails  ? 

Ans.  234qrs.  936n. 

6.  In  7248  nails,  how  many  yards,  ells  Flemish,  ells  English,  and 
ells  French  ? 

Ans.  453yds.  604  ells  Flem.  362  ells  Eng.  2qrs.  302  ells  French. 

7.  In  19  pieces  of  cloth,  each  15  yards,  2  quarters,  how  many 
yards,  quarters  and  nails  ?     Ans.  294yds.  2qrs.  1 178qrs.  and  4712n. 


6.  Long  Measure. 

1.  How  many  barley  corns  will  reach  fromNewburyportto  Bos- 
ton, it  being  43  miles  ? 
43  miles. 
8  3)8173440  proof.      Here  I  divide  by  11,  and 

multiply  the  quotient  by  2 
because  twice  5i  is  11 ;  or 
I  might  first  have  multipli- 
ed by  2,  and,  then,  have 
divided  the  product  by  11. 


344  furlongs, 
40 

12)2724480 
3)227040 
U)75680 

13760  rods. 

6880Q 
6880 

6880 
2 

75680  yards. 
3 

4 10)  137610 

8)344 

43 

227040  feet. 
12 

2724480  inches. 
3 

8173440  Answer. 

2.  How  many  barley  corns  will  reach  round  the  globe,  it  being 
360  degrees  ?  Ans.  4756801600. 

3.  How  many  inches  from  Newburyport  to  London,  it  being  2700 
miles  ?  Ans.  171072000. 

4.  How  often  will  a  wheel,  of  16  feet  and  6  inches  circumfer- 
ence, turn  round  in  the  distance  from  Newburyport  to  Cambridge, 
U  being  42  miles  ?  Ans.  13440  times. 

5.  In  190080  inches,  how  many  yards  and  leaguef^  ? 

Ans.  5280yds.  and  1  league. 


68 


REDUCTION. 

7.  Time. 

1.  In  20  years  how  many  seconds  ? 
(1.       h. 
365      6  in  a  ye^r. 
24 

1466 
730 


8766  hours  in  1  year. 

20 


176320  hours  in  20  years. 
60 


Proof. 

610)631  ]6200|O 

6|0)1051920|O 
2[0)17632|0 
4X6)8766 
4)1461 

365ci.  6h: 


10519200  minutes  in  ditto. 
60- 


631152000  seconds  in  ditto. 

2.  Suppose  your  age  to  be  15y.  19d.  llh.  37m.  45s.  how  many 
seconds  are  there  in  it,  allowing  365  days  and  6  hours  to  the  year  ? 

Ans.  475047465. 

3.  In  31536000  seconds  how  many  years?  Ans.   1  year. 

4.  How  many  minutes  from  the  first  day  of  January  to  the  14th 
day  of  August,  inclusively  ?  Ans.  325440. 

5.  How  many  days  since  the  commencement  of  the  ChristianiEra  ? 

6.  How  many  minutes  since  the  commencement  of  the  American 
war,  which  happened  on  the  19th  day  of  April,  1775? 

7.  How  many  seconds  between  the  commencement  of  the  war», 
April  19th,  1775,  and  the  independence  of  the  United  States  of  A- 
merica,  which  took  place  the  4th  day  of  July,  1776*  ? 

Ans.  38188800. 

8.  Motion. 

1.  In  9  si^ns,  13°  25*,  how  many  seconds  ? 

9s     13<»     25'                 *           6|0)102030|0  Proof. 
30  — ^ — : 


283  degrees. 
60 


6|0)1700|5 


3|0)2{ 


17005  minutes. 
60 


9s  13°  25' 


1020300  seconds. 


1776  was  a  leap  year. 


REDUCTION;  69 

9.  Land  or  Sgiuare  Measure. 

1.  In  29  acres,  3  roods,  19  poles,  how  many  roods  and  perches? 

Acres.  R.  Poles.  Proof. 

29     3     19  4|0)477|9 

— -  4)119~19p. 

119  roods.  • 

40  29ac.  3  roods. 

Answer  4779  perches. 

2.  In  1997  poles  how  many  acres  ?  Am.  12a.  Ir.  37p. 

3.  In  89763  square  yards  how  many  acres,  &c.  ? 

Ans.  18a.  2r.  7p.  lOlft.  36in. 

4.  How  many  square  feet,  square  yards,  and  square  poles,  in  a 
square  mile  ? 

Ans.  27878400  feet,  3097600  yards,  and  102100  poles. 

10.  Solid  Measure. 

1.  In  15  tons  of  hewn  timber  how  many  solid  inches  ? 
15  tons.  Proof. 

50  5|0 

—  1728)1296000(75|0 

750  feet.  1209a     

1728  15  tons, 

8640 

6000  ,  8640 

1500 
5250 
750 


Ans.   1296000  inches. 

2.  In  9  tons  of  round  timber  how  many  inches?  Ans.  622080. 

3.  In  25  cords  of  wood  how  many  inches  ?         Ans.  5529600. 
Grindstones  are  usually  sold  by  the  solid  foot,  and  the  contents 

are  found  by  the  following  Rule  ; — 

Multiply  the  sum  of  the  whole  diameter  and  of  the  half  of  the 
diameter,  by  the  half  diameter,  and  this  product  by  the  thickness, 
and  you  have  the  contents  in  cubic  inches. 

4.  What  is  the  content  of  a  grindstone,  whose  diameter  is  32 
inches  and  its  thickness  3  inches  ? 

32  diameter.  1728)2304(1  foot. 

16  half  diameter.  1728 

48  576 

16  3 

768  )1728(  1  third. 

3  thickness,  1723 


2304  solid  inches. 


Ans.   1  foot  and  ^  foot. 


# 


70  REDUCTION. 

5.  How  many  solid  feet  in  a  grindstone,  whose  diameter  is  40 
inches  and  thickness  4  inches  ?  Ans.  21  feet. 

Note.  This  rule  is  not  designed  to  give  the  solid  contents  with 
perfect  accuracy.     For  the  true  rule,  see  Mensuration,  Art.  30. 

11.  Wine  Measure, 

1.  In  9hhds.  ISgalls.  3qts.  of  wine  how  many  quarts? 

hhds.  gal.  qts.                                       Proof. 

9      15  3                                        4)2331 

63  

65)582— 3qt3. 

32  

65  9hhds — 15gals. 


582  gallons 
4 


Ans.    2331  quarts. 

2.  In  12  pipes  of  wine  how  many  pints  ?  Ans.  1209(J. 

3.  In  9758  pints  of  braudv  how  many  pipes  ? 

Ans.  9p.  Ihhd.  22gal.  3qt^ 

4.  In  1008  quarts  of  cyder  how  many  tons?  Ans.  1  ton, 

^         12.  Ale  or  Beer  Measure. 
1.  In  29hhds.  beer  how  many  pints  ? 

hhds.  Proof. 
29  2)12528 
54  

4)6264 


116 


14i?  54)1560 


15G6  galion:>  29  hhd*. 

4 


6264  quarts. 
,  2 


I 


Ans.    12528  pints. 

>.   In  47bar.  ISgal.  of  ale  how  many  pints?  Ans.  !3680i 

3.  In  36  puncheons  of  beer  how  many  butts?  Ans.  24. 

13.   Dry  Measure. 
I,  In  42  chaldrons  of  coals  how  many  pecks  ? 
Chaldrons.  Proof. 

42  4)5376 

32  32)1344(42 

"B4  1^> 

126  ^^4 

'l344  bushels.  I'l 

4 

Ans.  5376  peck?. 


"^-iw-- 


VULGAR  FRACTIONS.  71 

2.  In  75  bushels  of  corn  how  many  pints  ?  Ana*  4800. 

3.  In  9376  quarts  how  many  bushels  ?  Ans.  293. 


FRACTIONS. 

Parts  of  a  thing  are  expressed  by  figures,  as  well  as  whole  things. 
When  a  whole  is  expressed  by  figures,  the  number  is  called  an  in- 
teger. But  when  a  part,  or  some  parts  of  a  thing,  are  denoted  by 
figures,  as  one  fourth,  tu^o  thirds^  four  sevenths,  three  tenths,  &c.  of  a 
thing,  the  expressions  of  these  parts  by  figures  are  called  Fractions, 
The  term,  fraction,  is  derived  from  a  Latin  word,  which  signifies 
to  break,  as  an  integer  or  unity  is  supposed  to  be  broken  or  divided 
into  a  certain  number  of  equal  parts,  one  or  more  of  which  parts 
are  denoted  by  the  fraction.  Thus  one  fourth  denotes  one  of  the 
four  equal  parts,  and  three  tenths  denotes  three  of  the  ten  equal  parts, 
into  which  a  thing  is  broken  or  an  integer  divided. 

Fractions  arise  naturally  from  the  operations  of  Division,  when 
the  divisor  is  not  contained  a  certain  number  of  times  exactly  in 
the  dividend.  For  the  remainder  after  the  division  is  performed, 
is  a  part  of  the  dividend  which  has  not  been  divided ;  the  divisor 
being  the  number  of  parts  into  which  the  integer  is  divided,  and 
the  remainder- showing  the  number  of  those  parts  expressed  by  the 
fraction.  Thus  4  is  contained  in  9,  two  and  one  fourth  iimeSy  and, 
hence  the  quotient  cannot  be  folly  expressed  in  such  cases,  except 
by  a  whole  number  and  a  fraction. 

Fractions  are  divided  into  two  kinds,  Vulgar,  and  Decimal. 

VULGAR  FRACTIONS. 

Vnlgar  Fractions  are  expressions  for   any  assignable  parts  of  « 
unit,  or  whole  number  ;  and  are  represented  by  two  numbers  plac 
ed   one  above  another,  with  a  line   drawn   between  them,  thus  : 
f ,  |,  &c.  signifying  five  eighths,  four  thirds. 

The  figure  above  the  line  is  called  the  numerator,  and  that  beto^^ 
it  the  denominator. 

The  denominator  shews  how  many  parts  the  integer  is  divided 
into  ;  and  the  numerator  shews  how  many  of  those  parts  are  mean: 
by  the  fraction. 

Fractions  are  either  proper,  improper,  single,  compound,  or 
mixed. 

1.  A  single  or  simple  fraction  is  a  fraction  expressed  in  a  simple 
form  ;  as  a.  f ,  ^^,  &c. 

2.  A  compound  fraction  is  a  fraction  expressed  in  a  compound 
form,  being  a  fraction  of  a  fraction  ;  as  ^of-^,  f  of-\  of  if,  which 
are  read  thus,  one  half  of  three  fourths,  twosevenllis  of  five  elev 
enths  of  nineteen  twentielhs,  &c. 


72  VULGAR  FRACTIONS. 

3.  A  proper  fraction  is  a  fraction  whose  numerator  is  less  than 
its  denominator  ;  as  |,  |,  &c. 

4.  An  improper  traction  is  a  fraction,  whose  numerator  exceeds 
its  denominator  ;  as  |,  |,  &c. 

5.  A  mixed  number  is  composed  ofa  whole  number  and  a  fraction, 
as  7|,  35y''j,  &;c.  that  is,  seven  and  three  fifths,  &c. 

6.  A  fraction  is  said  to  be  in  its  least,  or  lowest  terms,  when  it 
is  expressed  by  the  least  numbers  possible. 

7.  The  common  measure  of  two,  or  more  numbers,  is  that  num- 
ber which  will  divide  each  of  them  without  a  remainder :  Thus,  5 
is  (he  common  measure  of  10,  20  and  30  ;  and  the  greatest  number, 
which  will  do  this,  is  called  the  greatest  common  measure. 

8.  A  number,  which  can  be  measured  by  two,  or  more  numbersj 
is  called  their  common  multiple  :  And,  if  it  be  the  least  number, 
which  can  be  so  measured,  it  is  called  the  least  common  multiple; 
thus,  40,  60,  80,  100,  are  multiples  of  4  and  5  :  but  their  least  com- 
mon multiple  is  20. 

Note.  The  product  of  two  or  more  numbers  is  a  common  mul- 
tiple of  those  numbers.  Thus,  3x4x5=60,  and  60,  or  3x4x5,  is 
evidently  divisible,  without  remainder,  by  each  of  those  numbers. 
a\nd  the  same  must  be  true  in  every  similar  case. 

9.  A  prime  number  is  one,  which  can  be  measured  only  by  itself 
or  a  unit,  as,  3,  7,  23,  &;c. 

10.  A  perfect  number  is  equal  to  the  Sum  of  all  its  aliquot  parts.* 
An  aliquot  part  ofa  number  is  contained  a  certain  number  of  times 
exactly  in  the  number. 

/  Problem  I.j 

To  find  the  greatest  common  measure  of  two^  or  2nore,  numhets. 

Rule. 
I.  If  there  be  two  numbers  only,  divide  the  greater  by  the  less, 
and  this  divisor  by  the  remainder,  and  so  on,  always  dividing  the 

*  The  foUowljig  perfect  numbers  are  all  which  are,  at  present,  known. 
6  85<S9869056 

28  137438691328 

496  5*305843008139952128 

8128  2417851639228158837784576 

33550336  9903520314282971830448816128 

f  This  and  the  following  problem  will  be  found  very  useful  in  the  doctrine 
of  fractions,  and  several  other  parts  of  Arithmetitk. 

The  truth  of  the  rule  may  be  shewn  from  the  first  example :  For,  since  io8 

measures  ai6,it  also  measures  cn6-j-io8,  or  324.  

Again,  since  108  measures  ai6  and  324,  it  also  measures  5  y  3244-216,  or 
1836.  In  the  same  manner  it  will  be  found  to  measure  axi836-}-324,  or 
3996,  and  so  on. 

It  is  also  the  grcatcsr  common  measure;  for  suppose  there  be  a  greater, then, 
since  the  greater  measures  1836  and  3996,  it  also  measures  the  remainder  324  *, 
and  since  it  measures  324  and  1836,  it  also  measures  the  remainder  216  ;  in  the 
same  manner  it  will  be  found  to  measure  the  remainder  108;  that  is,  the 
greater  measures  the  less,  which  is  absurd ;  therefore,  108  is  the  greatest  com- 
mon measure. 

In  the  same  manner;  the  demonstration  mny  be  applied  to  one  or  more  artl- 
djtional  nnmher.s. 


VULGAR  FRACTION^.  73 

last  divisor  by  the  last  remainder,  till  nolhing  remain,  then  will  the 
last  divisor  be  the  greatest  common  ra,easure  required. 
^  II.  When  there  are  more  than  two  numbers,  lind  the  greatest 
common  measure  of  two  of  them,  as  before  ;  then,  o{  that  common 
measure  and  one  of  the  other  numbers,  and  so  on,  through  all  the 
numbers,  to  the  last  ;  then  will  the  greatest  common  measure,  last 
found,  be  the  answer. 

in.  If  1  happens  to  be  the  common  measure,  the  given  numbers 
are  prime  to  each  other,  and  found  to  be  incommoaaurable,  or  '\tx 
Iheir  lowest  terms. 

Examples. 

1.  What  is  the  greatest  common  measure  of  1836,  3996,  and 
;044? 

1836)3996rs  So  108  is  the  greatest  common  measure 

3672^  of  39f)6  and  1836. 

Hence  108)1044(9 

324)1836(5  972 

16^>0  

72)108(1 

216)324'(1  72 

216  — 

Last  greatest  com.  meas.=36)72(2 

Common  meas.=108)21G(2  72 

216  — 

Therf  fore,  36  is  the  answer  required. 

2.  What  is  the  greatest  common  measure  of  1224  and  1080  ? 

Ans.  72. 
o'.  What  is  the  greatest  common  measAire  of  1440, 672  and  3472  ? 

Ans.  16. 
• 

Problem   II.* 

To  find  the  l^aat  ^mmon  multiple  of  tu^o  or  more  numbers. 

Rule. 

I.  Divide  hy  any  number  that  ^ill  divide  two,  or  more,  of  the 
ijiven  numbers  withnut  a  remainder,  and  set  the  quotients,  togethr 
er  with  the  undivi«led  numbers,  in  a  line  beneath. 

II.  Divide  the  second  line,  a©  before,  and  so  on,  till  there  are 
no  two  luimbers  that  can  t>e  divided;  Uien,  the  continued  product 
of  the  divisors  and  quotient*  will  give  the  multiple  required. 

*  The  reason  af  this  rule  nwyalso  l>e  shewn  froth  the  first  example:  Thu^, 
it  is  evident  that  6xioxi6x'JO  =  19100  triiy  be  divided  by  6,  lO,  16  ana 
20,  without  a  remainder  ;  hut  20  is  a  multiple  of  5  ;  therefore,  6x10X16x4, 
or  3840,  is  also  divisible  by  6,  10,  i6  and  20.  Also,  16  is  a  multiple  of  4; 
therefore  6x10x4X4-960,  is  ajso  divisible  by  6, 10, 16  and  to.  Also,  to  is 
a  multiple  of  2;  therefore,  6x5x4X4  -  480, is  also  divisible  by  6, 10,  i6and 
so.  Also,  6  is  a  multiple  of  2;  therefore,  3  X5  ^  4'<'4  -'*40,  is  also  divisible 
bv  6j  10,  i6,  ■luU  z&y  and  isWidcntly  the  least  iiuvober  fhat  can  t^c  59  divided* 

K 


^2)0 

2     16 

4 

^2)3 

1        8 

2 

*3 

1      *4 

1 

VULGAR  FRACTIONS. 

Examples. 

t.  What  is  the  least  common  multiple  of  6,  10,  16  and  20? 

I  survey  my  given  numbers,  and  find 
*5)6  10  16  20  that  five  will  divide  two  of  them,  viz. 
10  and  20,  which  I  divide  by  5,  bringing 
into  a  line  with  the  quotients  the  num- 
bers which  5  will  not  measure  :  Again, 
1  view  the  numbers  in  the  second  line, 
and  find  2  will  measure  them  all,  and 
get  3,  1,  8,  2  in  the  third  line,  and  find 
that  two  will  measure  8  and  2,  and  in 
the  fourth  line  get  3,  1,  4,  1,  all  prime  ; 
if-    *    *    *    *  I  ^[jgn  multiply  the  prime  numbers  and 

5x2x2x3X4=240  Ans.  the  divisors  continually  into  each  other, 
for  the  number  sought,  and  find  it  to  be 
240. 

2.  What  is  the  least  common  multiple  of  6  and  8  ?      Ans.  24. 

3.  What  is  the  least  number  that  3,  6,  8  and  10  will  measure  ? 

Ans.  120. 

4.  What  is  the  least  number  which  can  be  divided  by  the  9  di- 
gits, separately  without  a  remainder  ?  Ans.  2520. 

REDUCTION  OF  VULGAR  FRACTIONS 

Is  the  bringing  of  them  out  of  one  form  into  another,  in  order 
to  prepare  them  for  the  operations  of  Addition,  Subtraction,  &c. 

CASE  I.* 

To  abbreviate,  cr  reduce  fractions  to  their  lowest  terms. 

Rule. 

Divide  the  terms  of  the  given  fraction  by  any  number,  whicli 
will  divide  them  without  a  remainder,  and  the  quotients,  again,  in 

•  That  dividing  both  the  numerator  and  denominator  of  the  fraction  by 
the  same  number,  will  give  anottier  fraction  of  equal  value,  is  evident,  because 
both  patts  arc  diminished  proportionally,  and  if  both  parts  of  the  equal  frat- 
lion  be  multiplied  by  the  divisor,  the  original  fraction  will  be  formed  again. 

a88     8     36  36     8     288 

Thus    olH-T^T  ^^^^'Aq^'q^^'^''    ■^"'^  if  *hc  divisions  be  performed  as  of 

ten  as  can  be  done,  or  the  common  divisor  be  the  greatest  possible,  the  terms 
of  the  resulting  fraction  must  be  the  least  possible. 

Note  I.  Any  number,  ending  with  an  even  number  or  cypher,  is  divisible 
by  2. 

a.  Any  number,  ending  with  5  or  o,  is  divisible  by  5. 

3.  If  the  right  hand  place  of  any  number  be  o,  the  whole  is  divisible  by  ic. 

4.  If  the  two  right  hand  figures  of  any  number  be  divisible  by  4,  the  whole 
is  divisible  by  4. 

5.  If  the  three  right  hand  figures  of  any  number  be  divisible  by  8,  the  whole 
is  divisible  by  8. 

6.  If  the  sum  of  the  digits,  constituting  any  uumber,  be  divisible  by  3  or  9, 
the  whole  is  divisible  by  3  or  9. 


VULGAR  FRACTIONS,  75 

the  same  manner ;  and  so  on,  till  it  appears  that  there  is  no  nura- 
ber  greater  than  1,  which  will  divide  them,  and  the  fraction  will 
be  in  its  lowest  terms.     Or, 

Divide  both  the  terms  of  the  fraction  by  their  greatest  common 
measure,  and  the  quotients  vvill  be  the  terms  of  the  fraction  le- 
quired. 

Examples. 

1.  Reduce  f|f  to  its  lowest  terms. 

■        (4)    (3) 
«  lm=U=T%==i  the  answer.* 

Or  thus  : 
288)480(1  Therefore  96  is  the  greatest  commoii 

288  measure. 

and    96  J f If  =?  the  same  as  before. 

192)288(1 
192 

Com.  meas.  96)192(2 
192 

2.  Reduce  ^^\  to  its  lowest  terms.  Ans.  y\. 

3.  Reduce  y^sV  ^o  its  lowest  terms.  Ans.  ^, 

4.  Reduce  j^j\  to  its  lowest  terms.  Ans.  J  . 

7.  Tf  a  number  cannot  be  divided  by  some  number  less  tl^an  (he  square  root 
thereof,  that  number  is  a  prime. 

8-  A\\  prime  numbers,  except  a  and  5,  have  1, 3, 7, or  9  in  the  place  of  units  : 
and  all  other  numbers  are  compossite. 

9.  When  numbers,  with  the  sign  of  Addition  or  Subtraction  between  them, 
are  to  be  divided  by  any  numbers,  each  of  the  numbers  must  be  .divided  ; 
Thus  64-9-t-ia--a-f-3-(-4r-9,  or  6-f  9-f  ia=a7^9. 

3  3  T 

10.  But  if  the  numbers  have  the  sign  of  Mul..ipHcation  between  them  ;  then 

4x6x10    ax6xio      ax6x* 
only  one  of  tliem  must  be  divided  :  Thus,  — — t- — -= — 7 ~ — ■= — _  .   ,    :=^ 

axj  ^  ^^  S  AX* 

*  Hence  if  both  parts  of  a  fraction  be  multiplied  by  the  same  number,  it» 
,      ^      3     3     3      9       9      4     36     8     288 
value  IS  no  altered.     For -=7X7=,;=     X"— ,^X-?=7o-:,  and  so  on.     If 

5     b     i     T5      15     4     60     8     480 
fractions  be  multiplied  together,  in  which  equal  terms  occur  ia  the  numerator 
and  denominator,  these  equal  terms  may  be  cxpunnged  or  cancelled,  for  their 
quotient  would  be  i,  which  as  a  factor  would  not  alter  the  value  of  the  fraction, 

^^^      4     J     4X5     41  I3a4i3*4i  ^.. 

Thus-Xg'-^-^-g=-.3.--.  and   -x-x-X7:.-X^X jX-=7=i.  Anthmet- 

ic?I  operations  are  often  much  shortened  by  observing  what  quantities  may  be 
expunged, and  by  omitting  them  in  the  operations.     For  the  same  object,  ex- 
pressions may  be  changed  to  equivalent  ones,  and  quantities  expunged,     Thn? 
^     7>^     \    ^Xi5     I         8     a88     8    9x31    4  <  8     _8_ 
%  '^  45 "^^  ^3x15 ~ 3'*"^  9  ^  480^9  ^  8x6o=4Xi5"T^" 


7.0  VULGAR  FRACTIONS. 

5.  Reduce  y^i  ^^  *ls  lowest  terms.  Ans.  -]v 

6.  Reduce  ^|||  to  its  lowest  terms.  Ans.  v. 
»Yo^e.     If  the  numerator  of  a  fraction  be  multiplied,  or  it?  denom- 
inator divided,  by  a  whole  Dumber,  the  value  of  the  fraction  will 

1X3 
Im;  so  many  times  increased.     Thus,  i  multiplied  by  3.—    ('    = 

311 

-=-= _      Hence,  to   multiply  a  fraction  by  an  integer,  is  to 

y     fj     tj'  I  •  'O . 

multiply  the  numerator,  or  divide  the  denominator  of  the  fraction 

by  the  integer. 

1.  Multiply /j-  by  7.  Ans.  |f 

2.  Increase  the  value  of  yij,  nineteen  times.  Ans.  -^. 

3.  Increase  the  value  of  ^,  seven  fold.  Ans.  3. 
If  the  numerator  of  a   fraction  he  divided,   or   its   denominator 

multiplied,  by  a  whole  number,  the  value  of  the  fraction  will  be  so 

3-4-3     1        3 
many  times  diminished.     Thus,  f  divided  by  3,—-^- — =-=———-=; 

2T~9'  Henc<»,  to  divide  a  fraction  by  an  integer,  is  to  divide  the 
numerator,  or  multiply  the  denominator  of  the  fraction,  by  the 
whole  number. 

1.  Divide  f  by  7.  »       Ana.  ^%. 

2.  Diminish  the  value  of-},  seven  times.  Ans.  ^V- 

3.  Diminish  the  value  of  J,  four  times.  Ans.  j. 
Note.     The  reason  of  man"y  operations  will  be  evident  from  an 

attention  to  the  M\ovi'\n^  self -evident  truths. 

1.  If  equals  b«  added  to  equals,  their  sums  will  be  eq\ial.  Thus, 
34-44-9=8x2.  Let  7  be  added  to  each,  and  3+4+9+7=8x2! 
+7=23. 

2.  If  equals  be  subtracted  from  equals,  the  remainders  will  be 
equal.  Thus,  3+11=7x2.  Let  3  be  taken  from  each,  and  3+ 
11—3=7x2—3=11. 

3.  If  equals  be  multiplied  by  the  same  quantity,  the  products 
will  be  equal.  Thus,  let  5  +  7=6x2,  be  multiplied  by  6,  and 
.54:7x6=6x2x6=72. 

4.  If  equivalent  quantities  be  divided  by  the  same  quantity,  the 

quotients  will  be  equal,     'i'hus,  let  43+17=129<5  be  divided  by  6, 

43+17      12X5 

and  43+17-~5=12x5r^5,=  12,  or =—7— =12. 


CASE  II. 

To  reduce  a  mixed  number  to  its  equivalent  improper  fracti(jrh 

Rule.* 
Multiply  the  whole  number  by  the  denominator  of  the  fraction, 

*  All  fi'a(?llon9  rcprcfent  a  divifion  of  a  numerator  by  the  denominator,  and 
are  taken  altojirtlier  as  proper  and  adequate  cxprcflions  of  the  quotient.  Thus 
tlie  quotient  of  3,  divided  by4,is-|;  from  whence  the  rule  is  manifcft ;  forif 
any  number  is  multiplied  and  divided  by  the  fame  number,  i>is  cvidtnt  ihf 
,tjfuotitut  nuift  be  ihc  fame  as  the  quantity  firft  given. 


VULGAR  FRACTIONS.  77 

and  add  the  numerator  of  the  fraction  to  the  product ;  under  which 
euhjoin  the  denominator,  and  it  will  form  the  fraction  require*!. 

Examples. 

h  Reduce  36f  to  its  equivalent  improper  fraction. 

36  I  multiply  36  by  8,  and  adding  the  nu- 

X8-{-5  merator  5  to  the  product,  as  I  mul- 

tiply,  the  sum  293  is  the  numerator^ 

Ans.  293  of  the  fraction  sought,  and  8  the  de* 

nominator:     So  that  2 13  i^  the  im- 

8  proper  fraction,  equal  to  36f. 

36x8+5=293  Answer  as  before. 
Or,         g  ~8~ 

2.  Reduce  127 j\  to  its  equivalent  improper  fraction.     Ans.  ^{^',. 

3.  Reduce  653,^^^  to  its  equivalent  improper  fraction. 

^  Ans.  *2^«». 

CASE  Ill.t 

To  reduce  a  tzhole  number  to  an  equivalent  fraction  having  a  given 
denominator. 

Rule. 

Multiply  the  whole  number  by  the  given  denominators 
Place  the  product  over  the  said  denominator,  and  it  will  form  the 
fraction  required. 

Examples. 

1.  Reduce  6  to  a  fraction,  whose  denominator  shall  be  8. 

6X8=48,  and  V  the  Ans — Proof  V  =48-^-8=6. 

2.  Reduce  15  to  a  fraction,  whose  denominator  Shall  be  12. 

Ans.  V^^ 

3.  Reduce  100  to  a  fraction,  whose  denominator  shall  be  70. 

Ans.  '^^^''=»|o=100. 

A  whole  number  is  made  a  fraction  by  drawing  a  line  under  it, 

9x1 
and  putting  unity  or  1,  for  a  denominator,  as  r=— j — by  therule, 

^nd  12  i.s  V^*c> 

CASE  IV.J 

To  reduce  an  improper  fraction  to  its  equivalent  whole,  or  mixed 

number. 

Rule. 
Divide  the  numerator  by  the  denominator:  the  quotient  will  b« 
the  whole  number,  and  the  remainder,  if  any,  will  be  the  numera- 
tor to  the  given  denominator. 

f  Multiplication  and  Divifion  arc  here  equally  ufed,  and  confequently  the  re- 
sult is  the  fame  as  the  quantity  firft  propofcd. 

i  This  cafe  is,  evidently,  the  revcrfc  of  qafc  ^d,  and  his  its  reafon  m  the  na- 
ture •f  €«ram«Bi  divifioB. 


78  VULGAR  FRACTIONS. 


1.  Reduce  2|3  to  its  equivalent  whole,  or  mixed  number. 

8)293(36|-  Ans. 
24 

53 
48 

—  Or,  2|3— 293-~8=36f  as  before. 

5 

2.  Reduce  ^ijs  to  its  equivalent  whole,  or  mixed  number. 

Ans.  127^t, 
3»  Reduce  *^^9^°  to  its  equivalent  whole,  or  mixed  number. 

Ans.  653/^. 
4.  Reduce  Y  to  its  equivalent  whole  number.  Ans.  9. 

CASE  v.* 

To  reduce  a  compound  fraction  to  an  equivalent  simple  one. 

Rule. 

Multiply  all  the  numerators  continually  together  for  a  new  na- 
me rator,  and  all  the  denominators,  for  a  new  denominator,  and  they 
will  form  (he  simple  fraction  required. 

If  part  of  the  compound  fraction  be  a  whole  or  mixed  number,  it 
must  be  reduced  to  an  improper  fraction,  by  case  2d,  or  3d. 

if  the  denominator  of  any  member  of  a  compound  fraction  be 
equal  to  the  numerator  of  another  member  thereof,  these  equal 
numerators  and  denominators  may  be  expunged,  and  the  other 
members  continually  multi{)iied,  as  by  the  rule,  will  produce  the 
iractioDS  required  in  lower  terms. 


E 


XAxMPLES. 


1.  Reduce  -I  of  |  of  f  off  to  a  simple  fraction. 

1X2x3x4      24       1 

*7. — z — : — r=T7i7r=7  the  Answer. 
2x3x4X5     120     5 

Or,  by  expunxiiig   the  equal  numerators  and  denominators,  it 

will  give  ]  as  before. 

2.  Reduce  |  of  5  off  of  |^  to  a  simple  fraction. 
3X4X5X11       660       n  • 
4y^/Jy>d2'^T44i)""24  '^"''     ^'*'  ^'^  expungmg  the  equal  nu- 

3X1 1 

^jeratr>rs  and  denominators,  it  will  be  =5b=^i  2is  before. 

6X12     ^^      ^* 

*  TWil  a  conipouud  fradlJonmay  be  reprefcnted  by  a  fimple  one  is  very  cvi- 
u'iit ;  fincc  a  part  of  a  part  muft  be  equal  to  fome  part  of  the  whole.  The 
ruth  of  the  rule  for  this  redudlion  may  he  fliown  as  follows. 

l,et  the  compound  fra(5lion  to  be  reduced,  be  1  of  _«_.  Then  |  of  _6_=_6_ 
t.  i_=:  Q,  and  confequentlv -5  of    r,  =A.vS=-li?  the  lame  as  by  tie  rule. 

^  -_  0  .  ^  '     i  IT)  4  0  -^  4  0  ^  . 

i{  the  compound  fratflion  confifts  of  more  numbers  than  two,  the  firft  two 
raay  be  reduced  to  one,  and  that  one  and  the  tliird  will  be  the  fame  as  a  frac- 
■'x>n  of  two  Bunibers,  and  fo  oi». 


VULGAR  FRACTIOxXS.  7S 

3.  Reduce  f  of  ^  of  |f  to  a  a  simple  fraction.  Ans.  ||f. 

4.  Reduce  j\  of  H  of  ,^  of  20  to  a  simple  fraction. 

Ans    3  2±=z9_2_ 

^"^'    30«  51- 

5.  Reduce  i  of  ^  of  |  of  12^  to  a  simple  fraction.     Ans.  U^Hh 

CASE.  VI. 

To  reduce  fractions  of  different  denominators  to  equivalent  fractions 
having  a  common  denominator. 

Rule  I.* 

Multiply  each  numerator  into  all  the  denominators  except  its 
own,  for  a  new  numerator,  and  all  the  denominators  into  each  oth- 
^r,  continually,  for  a  common  denominator. 

Examples. 

1.  Reduce  a,  f ,  and  f  to  equivalent  fractions  having  a  common 
denominator.  1X5 x  8=40  the  new  numerator  for  |. 

~X4x8=  64  the  new  numerator  for  |. 
5x4x5=100  ditto  for  |. 

4x5x8=160  the  common  denominator. 

Therefore  the  new  equivalent  fractions  are  -j^o*  tVo  ^^^  lih 
the  answer. 

2.  Reduce  |,  |,  f,  |^  and  {-  to  fractions  having  a  common  denoDj- 

inafnr  "  Ans    -fJ-^-,  -'L6_8_    jlbjl    JUUL    l«i>8 

indlOr.  /ins.    1^X5  2  >    1152'     IIS2»    1T52»    Tl52* 

3.  Reduce  ^,  |  of  |,  7|,  and  j\,  to  a  common  denominator. 

Ana        (>3C         1040       14503        432 

^^^-  T8T2»  T»?2'     T8T"2  >  TsTi 

4.  Reduce  |],  f  of  2i,  y'j,  and  |-,  to  a  common  denominator. 

Ans      -"AO-      216  0  0.     JL72JL     _T3  0  0 
•""''•     ild20»     11J20J    11S20>    IIoIt>' 

Rule  H. 

To  reduce  any  given  fractions  to  others^  which  shall  have  the  least 
common  denominator. 

1.  By  Problems,  Page  73,  find  the  least  common  multiple  of  all 
the  denominators  of  the  given  fractions,  and  it  will  be  th€  common 
denominator  required. 

2.  Divide  the  common  denominator  by  the  denominator  of  eacL 
fraction,  and  multiply  the  quotient  by  the  numerator,  and  the  pro- 
duct will  be  the  numerator  of  the  fraction  required. 

*  By  placing  the  numbers  multiplied,  properly  under  one  another,  it  will  be 
seen  that  the  numerator  and  denominator  of  every  fraction  are  multiplied  by 
the  very  same  number,  and  consequently  their  values  are  not  altered.  Thus, 
m  the  first  example. 


1|X5X8 


4|x5xa 

In  the  second 


X4X! 


X4XH 


X4X5 


X4X5 
rule,  tlie  common  denominator  is  a  multiple  of  all  the  denom- 
inators, and  consctjucntly  will  divide  by  any  of  them  ;  Therefore,  proper  part« 
may  be  taken  for  aiK  the  numerators  as  required. 


80  VULGAR  FRACTIbNS. 

Examples. 

1.  Reduce  ^,  ^i  and  |-  to  fractions  having  the  least  common  de 
nominator  possible. 

4)3     4     8  4X3X2=24=  least  common  de- 

nominator. 

3     1     2 


24-r-3xl=8  the  first  numerator;  24-~-4X3=l^  the  second  uu« 
nierator;  24-^8x7=21  the  third  numerator. 

Whence,  the  required  fractions  are  gl^,  If,  §}. 

2.  Reduce  J,  |,  f,  and  4  to  fractions  having  the  least  common 
denominator.  "    "  Ans.  ^a,  aj,  ia,  and  |^. 

CASE  VII. 

To  reduce  a  fraction  of  one  denomination  to  an  equivalent  fraction  of 
a  higher  denomination. 

Rule.*- 

Blultiply  the  given  denominator  by  the  parts  in  the  several  i\c^ 
nominations  between  it  and  that  denomination  to  which  it  is  to  be 
reduced,  for  a  new  denominator,  which  is  to  be  placed  under  the 
given  numerator  :  Or,  compare  the  given  fraction  with  the  several 
denominations  between  it  and  that  denomination  to  which  it  is  to  be 
reduced,  and  then,  by  case  6th,  reduce  the  compound  fraction  thus 
formed,  loa  single  one,  and  the  equivalent  fraction  of  the  required 
denomination  will  be  obtained.  Let  this  fraction  be  reduced  to  its 
lowest  terms. 

*  The  reafon  of  the  rule  may  be  foea  in  the  following  manner.  As  there  arc 
la  pence  in  a  fhilling,  foiw-fifths  of  one  penny  can  be  only  a  t-welfth  part  as 
much  of  la  pence  or  a  fliilling,as  it  is  of  one  penny.  Hence,  to  reduce  four 
fifths  of  a  penny  to  the  fraAion  of  a  fliilling,  the  given  fradlion  muft  be  di- 
miniflied  la  times,  or  one  twelfth  of  it  will  be  the  equivalent  fradlion  of  a 
fhilling  A  fradlion  is  diminiflied  in  value,  according  to  the  note  to  Cafe  1.  by 
multiplying  the  denominator  by  the  whole  number.    Thus  four  fifths  of  a  pen- 

4  4       14        14 

ny  =r  - — — -of  afliil!ing=:— y—  =—  of  — =:  —  of  a  fliilling.     For  the  fame  rea-- 

fon,fow  fixticths  of  a  fliilling  can  be  only  one  twentieth  a^  much  of  a  pound, 

pound.     Put  thefe  two  operations  together,  and  you  have  four-fifths  of  a  penny. 

The  fam«  operation  might  have  been  performed  thus.     In  a  pound  there  are 

4  4  1 

210  pence.     Then,  four-fifths  of  a  penny  = ,—  of  a  pound,  ==  -  of  — :~k 

'       5X'^40  a         240 

-- -  as  before.     And  in  general  the  fradkion  of  one  denomination  muft  be  as 

much  diminifliedto  be  an  equivalent  fra<flion  of  a  higher  denomination,  as  is 
indicated  by  the  numS^cr  of  patts  of  t.hc  givcn  denomination  to  make  •ne  of  the 
tijffher  derwfvihiat'ir'Q. 


VULGAR  FRACTIONS  81 

EXA^IPLES. 

I.  Reduce  f  of  a  cent  to  ttie  fpaction  of  a  dollar. 

By  comparing  it,  it  becomes  |  of  J^  of  J^,  which,  reduced  by 
case  5,  will  be  4X1  Xl  =    4 

~^Th  D.  An.. 
and  7x10x10=  700 

9.  Reduce  |  of  a  mill  to  the  fraction  of  an  eaijlo.  Ans.  t-^-tt  ^" 
ri.  Reduce  \\  of  a  mill  to  the  fraction  of  a  dollar.  Ans.  vtt~  D. 
4.  Reduce  |  ot  a  pennj  to  the  Iraction  of  a  pound.  Ans.  i^j^^. 
£».  Reduce  ^  of  a  farthing  to  the  fraction  of  a  pound.  Ans.  y^VV- 

6.  Reduce  |  of  a  penny  to  the  fraction  of  a  guinea. 

^"s.  j^\j  guinea. 

7.  Reduce  j|  of  a  shilling  to  the  fraction  of  a  moidore. 

Ans.  -Jj  moidore. 
€.   Reduce  4  of  an  ounce  to  the  fraction  of  a  jfe.  Avoirdupois. 

Ans.  -^\m. 
^9.  Reduce  ^  of  a  pound  to  the  fraction  of  a  guinea.  Ans.  ^guin. 

10.  Retluce  {■  of  a  pwt.  to  the  fraction  of  a  pound  Troy. 

Ans.  To'^ao^' 

II.  Reduce  -^  of  a  Ih.  Avoirdupois  to  the  fraction  of  1  Cut. 

Ans.  yJ-^  Cwt. 
12.  Express  &*  furlongs  in  the  fraction  of  a  mile.     Ans.  }i  mile. 

CASE  VIll. 

7t)  reduce  a  fraction  of  one  denomination  to  an  equivalent  fraction 
of  a  lower  denomination. 

liULE.t 

MuUiply  the  given  numerator  by  the  parts  in  the  denominations 
betiveen  it  and  that  denomination   you   would  reduce  it  to,  for   a 

4        4        20      CO  80  1        80      4 

f  This  rule  is  the  reverse  of  the  prercding,  and  the  propriety  of  it  may  be 
seen  in  a  similar  manner.  The  fraction  of  a  higher  denomination  is  obviously 
less  than  the  equivalent  fraction  of  a  lower  denomination  ;  for  instance,  ^  of 
a  pound  is  A  shillings  or  5  shillings.  Whence  the  value  of  the  fraction  muse 
te  increased,  to  render  ir  an  equivalent  frartion  of  a  lower  denomination,  so 
many  times  :hs  there  are  parts  of  the  less  denomination  in  the  higher.  But, by 
the  Note  to  Case  /,  the  value  of  a  fraction  is  increased  by  multiplying  the  nu- 

I 
merd^tor  by  a  whole  number.     To  reduce  £  to  the  fraction  of  a  shilling, 

■'  400 

I  ao 

as  there  are  20  shillings  in  a  pound,  we  Jiuve""""  Xa9="VQ-  of  a  shilling.  And 

to  reduce  ~-  of  a  shilling  to  the fi action  of  a  penny,  we  have    7qq^^^~^qo 

of  a  penny  =— d.     Put  together  these  operaiiens, and  we  have  ^^y-^  ^^4PO  -'' 

aoxi2v0f  a  pcrtuy  tni -;"  oX  --  of  " -'::x~ - --.-^^1,  as  before.' "-A-hJch  is  tlie 
^w  *  i      .43-1'     < 


82  VULGAR  FRACTIONS. 

new  numerator,  which  place  over  the  given  denominator  :  Or,  on-* 
]y  invert  the  parts  contained  in  the  integer,  and  make  of  them  a 
compound  rraction  as  before,  then,  reduce  it  to  a  simple  one. 

Examples. 

1     Reduce  jl-^  of  a  dollar  to  the  fraction  of  a  cent. 
By  comparing   the  fraction    it  will   be   yJ-j  of  Y  of   Y  ;  then 
1    *x    10  X  10        100        4 
m  X  T  X  T  =  "175  =  ?  ^-  ^"^^'*^^- 

2.  Reduce^o^o-g  of  an  eagle  to  the  fraction  of  a  mill.     Ans.  |m. 

3.  Reduce  T5V00  <^^'  ^  dollar  to  the  fraction  of  a  mill.     Ans.  |im. 

4.  Reduce  ^^^  ^'^^  pound  to  the  fraction  of  a  penny.       Ans.  |d. 

5.  Reduce  yaVo  *^^^  pound  to  the  fraction  of  a  farthing.    Ans.  |qr. 

6.  Reduce  o/jg^  of  a  guinea  to  the  fraction  of  a  penny.    Ans.  |d. 

7.  Reduce  j\  of  a  moidore  to  the  fraction  of  a  shilling.  Ans.  ||s. 

8.  Reduce  -^^  of  a  Ife  Avoirdupois  to  the  fraction  of  an  ounce. 

Ans.  ^oz. 

9.  Reduce  ^  of  a  guinea  to  the  fraction  of  a  pound.  Ans.  |£, 

10.  Reduce  ^/g^^  of  a  ife  Troy  to  the  fraction  of  a  pwt.   Ans.  fpwt. 

11.  Reduce  y^g  of  Cwt.  to  the  fraction  of  a  Ife  Avoirdupois. 

Ans.  fife. 

CASE  IX. 

To  find  the  value  of  a  fraction  in  the  known  parts  of  the  integer,  as 
of  coin i  weight,  measure,  4*c. 

Rule.* 
Multiply  the  numerator  by  the  parts  of  the  next  inferior  denomi- 
nation, and  divide  the  product  by  the  denominator  ;  and  if  any  thing 

4 
*  This  rule  follows  from  the  preceding.      Thus  let  ~£,  be  the  fraction, 

4         4         20 
whose  value  is  to  be  found.    By  the  preceding  rule, "TjS  =^r-  of  —-   of  a  shil- 

80  .     ^  20  .   .          40 

ling,- --^  s.  =  16s.     Aguin,j£=  j  of  —  of  a  shillings  ^s.  ==    by   dividing, 

12  12 

Ijjs.  And  on  the  &amt  principle,  |s.=:  |-  of  ~  of  a  penny,=:—  d.  =  4d.  Whence 

|X  =  1-|3.  =:   13s.    Id. 

J         3       '20  00  4  4  4  12 

Again,  YJ£—,y  of  j- of  a  shilling,  ^ys.  =  8~s.    Butrs.  =  y  of  —of  a 

48  6  4  r>  G         6         4 

penny, =^d.=  67;  d.  and,  therefore  [r-s.  =8s.C-;;;d.  But  -d.=--  of  7  of 
*        ^         /  <  7  V  til 

21  ..  3  4  0  3 

a  farthing,  =  --  qr.  =  3yqr.     Tliercfore,  7^£  —  8-9.  =  8s.  G-d.— .88.6d.:i7qr. 

The  same  process  is  obviously  applicable  to  every  similar  case.     Or,  the  process 

12         4      2880  :> 

may  be  conducted  thus;   ~X=^-':7  of —- of  —  of —=;—;:;- qr.  ^=  411— qrs.  ==: 

102d.  oTqrs.  --  Cs.  Od.  ::;rqrs. 


VULGAR  FRAeTIQNS. 


m 


vemain,  multiply  it  by  the  next  inferior  denomination,  and  divide 
by  the  denominator  as  before,  and  so  on,  as  far  as  necessary  ;  and 
the  quotients  placed  after  one  another,  in  their  order,  wUl  be  the 
answer  required  ;  or,  reduce  the  numerator,  as  if  it  were  a  whole 
number,  to  the  lowest  denomination,  and  divide  the  result  by  the 
denominator  ;  the  quotient  will  be  the  number  of  the  lowest  denom- 
ination, (which  must  be  .brought  into  higher  denominations  as  far  as 
it  will  go,)  and  the  remainder  will  be  a  numerator  to  be  placed 
overthe  given  denominator  for  a  fraction  of  the  lowest  denomination. 
Note.  From  this  rule,  in  connexion  with  what  has  been  said 
o[  Reduction  of  Federal  Money,  it  appears,  that,  annexing  to  the  giv- 
en numerator  as  many  cyphers,  as  wjH  fill  all  the  places  to  the  low- 
est denomination,  and  dividing  the  number  so  formed  by  the  de- 
nominator, the  quotient  will  be  the  answer  in  the  several  denomi- 
nations, and  the  remainder  a  numerator  to  be  placed  overthe  given 
denominator,  forming  a  fraction  of  the  lowest  denomination. 


Examples. 

1.  What  is  the  value  off  of  a  dollar  ? 
By  the  general  rule. 
5 
10 


By  the  note. 
$    d.    c.    m. 
8)  5     0     0     0 


8)50(     2 

$e    10 


Ans.  6d.  2c.  5m. 
8)20(  4  or  62c.  5m. 
c,  2   10 


6     2     5 


8)40 

m.  5 

Or  thus. 

j^5=5000m.     and  « o_pom.=:625m.=62c.  5m. 
2.  What  is  the  value  of  ^  of  a  dollar? 
g     d.    c.    m. 

64)17     0     0     0 


Ans,  as  before. 


128 

420 
384 


360 
320 


(2d.  6c.  5|m. 


or  26c.  5f  m.  Ans. 


Or,  ^17=17000ra.     And 


»V7"m.=265fm.^ 


40 


64 


$6c.  5|m. 


Ans.  as  before. 


J,  What  is  the  value  of/,-  of  an  eagle  ?  Ans.  $1  87c.  5ra. 


Sd  VULGAR  FKACTiONS. 

4.  What  is  Uie  value  of  y"^  of  a  dollar  ?  Atis.  4, 


5.  What  is  the  value  ot  4  ^f  a  pound  ?  Ans.  14s.  3d.  Ifqr. 

(5.  What  is  the  value  of  ^^^  ^^^  shilling?  Ans.  4id. 

7.  What  is  the  value  o\\\  of  a  £  ?  Ans.  ^.  Gt]. 

8.  What  is  the  value  of  f|  of  a  pistole  ?  Ans.  13s.  6d. 
IK  What  is  the  value  of  J.^  of  a  Cwt.  ? 

Ans.  2  qrs.  9fe  10  oz.  7|idi% 
10.  What  is  the  value  of|  of  a  ife  Avoirdupois? 

Ans.  12oz.  12|dr. 
n.  W^hat  is  the  value  off  of  a  ife  Troy  ?  Ans.  7oz.  4pwt. 

.12.   What  is  the  value  of  ^3  of  a  ton  ? 

Ans.  4cvvt.  Sqrs.  12Ife.  14oz.  12y\dr. 
13.  What  is  the  value  of  f  of  a  yard  ?  Ans.  2qrs.  2|n. 

M.   What  is  the  value  of  i  of  an  ell  English  ?         Ans.  4qrs.  l|n. 

15.  W'hat  is  the  value  off  of  a  mile  ?  Ans.  6fur.  26[>.  Hi]. 

16.  What  is  the  value  oi\%  of  a  day  ?       Ans.  16h.  36m.  65yy. 

17.  The  value  of -}|  of  a  Julian  year  is  required  ? 

Ans.  257d.  i9h.  -I5m.  52ffs. 

18.  'i'he  value  of  y\  of  a  guinea  h  demanded  h-j  Ans-  18.-;. 

19.  What  is  the  value  of  jS-ofa  dollar.  '  Ans.  6s.  7|d. 
520.  What  is  the  value  of  f  of  a  moidore  ?  Ans.  21s.  ^^. 
21.   What  is  the  value  of  5-  of  an  acre  ?                    Ans.  3r.  l/'/p. 

CASE  X. 

To  reduce  any  givc7i  quantily  io  the  fraction,  of  any  greater  iJenomi^ 
nation  of  the  same  kind. 

Reduce  the  given  quantity  to  the  lowest  term  mentioned,  for  a 
nuuieralpr;  then  reduce  the  integral  part  to  the  same  term  for  <♦ 
detiominator  ;  which  wiil  be  the  IVaclion  required- 
Note.  It  appears  fiom  this  rule  and  what  has  been  said  boforc, 
that,. in  Federal  Money,  where  the  given  quantily  contains  no  frac- 
tion of  its  lowest  denomination,  the  annexing  of  as  many  cyphers  {q 
i  of  ihe  required  denojnination,  as  will  extend  to  the  lowest  denoni.. 
ination  in  the  given  quantily,  will  forma  denominator,  which  plac- 
ed under  the  given  quantity  used  as  one  number  for  a  numerator, 
%vi!l  make  the  answer,  which  may  be  reduced  to  its;  lowest  term-j. 
Or,  if  there  be  a  fraction  of  (lie  lowest  denomination,  multiply  the 
given  whole  numbers  by  its  denominator,  adding  its  numerator,  for 
a  numerator  ;  and  let  the  denominator  itself  at  ihe  left  of  as  many 
cyphers  a^  were  mentioned  above  be  a  denominator  ;  the  fraction 
so  f)rmed  will  be  the  answer;  winch  may  be J^educed  to  its  lowest 
terms. 


*  This  cafe  in  tlic  rcvcrfc  of  llic  furnKT,  and  tlie  proof  evident  from  that. 
Note.     If  there  he  a  fraction  jiivcn  wlt)i  the  fjiid  quaDflty,it  mufl  he  fartiic. 
TcJiiud  to  the  dcnon.iuative  partb  thereof,  adding  iLtreto  the  numcratqr. 


VULGAR  FRACTIONS.  85 

Examples. 

1.  Reduce  6d.  2c.  6m.  to  the  fraction  of  a  dollar. 
Bv  the  general  rule. 
Cd.  "  lOd.  int.  pt. 

XlO+2  10 

(32  100  By  the  note. 

Xl0-f5  10  $    d.    c.    m. 

6     2     6 

625  1000  =1$. 

10     0     0 
And,    TVoV=t$  Ans.  Ans.  as  before. 

'i,'.  Reduce  26c.  5|«i.  to  the  fraction  of  a  dollar. 

By  the  general  rule.  By  the  note. 


26c.  iOOc.  Int.  pt.  g  d.  c.  m. 

XlO+5  10  265x84-5=^   1  2  5 


265  1000  And  §1x8=8  0  0  0 

X84-5  8 


=m- 


Ans.  as  before. 


2125  8000 

And,fie=HSAns. 

3.  Reduce  $1  87c.  5m.  to  the  fraction  of  an  eagle.       Ans.  ^j"^-. 

4.  Reduce  43c.  l^m.  to  the  fraction  of  a  dollar.  Ans.  y'^^. 

5.  Reduce  143.  Sid.^  to  the  fraction  of  a  pound.  Ans.m^=f£. 

6.  Reduce  4|d.  to  the  fraction  of  a  shilling.  Ans.  |%. 

7.  Reduce  3s.  Gd.  to  the  fraction  of  a  pound.  Ans.  -^q£, 

8.  Reduce  I3s.  6d.  to  the  fraction  of  a  pistole.      Ans.  ||pistole. 

9.  Reduce  2qrs.  9fe.  lOoz.  7|4dr.  to  the  fraction  of  a  cwt. 

Ans.  ||cwt. 
iO.  ReJuce  12oz.  12|dr.  to  the  fraction  of  a  ife  Avoirdupois. 

Ans.  |-ife. 
11.  Reduce  7oz.  4pwt.  to  the  fraction  of  a  Ife  Troy.  Ans  |lb. 
^2.  Reduce  4cwt.  2qr3.  12ife  14oz.  12y«^dr.  to  the  fraction  ofaton^ 

Ans.  vs^on. 

13.  Reduce  2qrs.  2?jn.  to  the  fraction  of  a  yard.  Ans.  fyd. 

14.  Reduce  4qra.  llu.  to  the  fraction  of  an  ell  English. 

Ans.  |E.  E. 

15.  Reduce  6fur.  26po.  lift,  to  the  fraction  of  a  mile.     Ans.  fm. 

16.  lieduce  16b.  36m.  55/3S.to  the  fraction  of  a  day.  Ans.  j\day. 

17.  Reduce  257d.  19h.  45in   52}-«s.  to  the  fraction  of  a  Julian  year, 

Ans.  If  J.  y, 
1^.  Reduce  I83.  to  the  fraction  of  a  guinea.  Ans.  Yi^^- 

19.  Keduce  5s.  T^d  to  the  fraction  ol' a  dollar.  Ans.  y|dol. 

20.  Reduce  21s.  Tid.  to  the  fraction  of  amoidore. 

Answer  fmoidorc. 

21.  Reduce  3r.  17:r>.  to  the  fraction  of  an  acre.  An:-,  facre^ 


S^  ^        VULGAR  FRACTIONS. 

ADDITION  OF  VULGAR  FRACTIONS. 

Rule.*  v 

Reduce  compound  fractions  to  single  ones ;  mixed  numbers  to 
improper  fractions ;  fractions  of  difterent  integers  to  those  of  the 
same  ;  and  all  of  them  to  a  common  denominator ;  then  the  sum 
of  the  numerators  written  over  the  common  denominator  will  be 
the  sum  of  the  fractions  required. 

Examples. 
1.  Add  7|,  -^-  off,  and  7  together. 
Flr-st       74  =  s_9    5  of  ^=iA    and  7=^ 

Then  the  fractions  are  ^j  II»  ^^^  t  >  therefore, 

39x56x   1=2184 


15X  5X   1=      75 
7X  6X56=1960 


4219  Or  thus, 


2184-1-75+1960 


280 


— — —  1  ^  1  ^ 
5x56x1=280 

2.  Add  -f,  91,  and  |  of  i  together.  Ans.  ^^\. 

3.  What  is  the  sum  of  |,  f  of  |  of  i,  and  8/3  ?         Ans.  9^2^^. 

4.  What  is  the  sum  of  /^  of  4f ,  f  of  i,  and  91  ?  Ans.  12ff. 

5.  Add  together  fE.  |g  and  lie.  Ans.  p  53c.  24m. 
f>.  Add  together  ij  fc.  ^^^  c.  and  ^ra.  Ans.  20c.  9m. 
7.  Add  £1  -fs.  and  fd.  together.  Ans.  2s.  8yV_d. 
Z.  What  is  the  sum  off  of  £17,  £9f  and  I  of  1  of  £A? 

Ans.  £16  12s.  3-fd. 
9.  Adr    I  of  a  yard,  1  of  a  foot,  and  f  of  a  mile  together. 

Ans.  1100yds.  2ft.  7inche<?. 

10.  Add  1  of  a  week,  1  of  a  day,  |  of  an  hour,  and  f  of  a  minute 

Ic^gcther.  Ans.  2  days,  2  hours,  30  minutes,  45  seconds. 

SUBTRACTION  OF  P^ULGAR  FRACTIONS. 

RuLE.j 

Prepare  the  fractions  as  in  Addition,  and  the  difference  of  the  nu- 
merators, wriifen  above  the  common  denominator,  will  give  the 
tliiference  of  the  fractions  required. 

*  Fradlions, before  they  are  reduced  to  a  common  denominator,  are  entirely 
ui lumllar, and  tlierefoie  cannot  be  incorporated  with  one  another;  but  when 
they  are  reduceJ  to  a  common  denominator,  and  made  parts  of  the  fame  thing  ; 
their  fum,  or  difference, may  then  be  as  properly  cxprefTed  by  the  fnm  or  dif- 
Icrcnce  of  the  numerators,  as  the  fum  or  dtfFerence  of  any  two  quantities  what- 
'cver,  by  the  fum  or  difference  of  their  individuals  ;  whence  the  rcafon  of  the 
tales,  l)oth  for  Addition  and  Subtradlion,  is  manifeft.  If  the  given  fractions 
f,ave  the  fame  denominator  and  are  of  the  fame  denomination,  the  fum  of  the 
numerators  written  over  the  given  denominator,  will  be  the  fum  of  the  fractions. 

f  In  fubtradling  mixed  numbers,  when  th-e  fradllons  have  a  common  denorai- 
D3tor,  and  the  numtratoi  in  the  fubtrahend  is  lefs  than  that  in  the  minuend, 
the  difference  of  the  whole  numbers  will  be  a  whole  number,  and  the  difference 
«f  the  numerators  a  numerator  to  be  placed  over  the  given  denominator;  tbh 


VULGAR  FRACTIONS.  87 

Examples. 


1.  From  f  take  f  of 


_  of  f=i|=^V     Then  the  fractions  are  f  and-\. 
3X28=     84  )  s=,8_4_  and -''-=---^-  Iherpforp 

^A    t  >i.u  ^      3J1..___2_0_=::_6J_=:4    rpmainrlpr 

4X28  =  112  com.  den.         J  112  112     112     7 

2.  From  ff  take  f.  .^W5.  if^, 

3.  From  371  take  19f  .^ns.  n^f. 

4.  From  13i  take  |  of  15.  Ans,  2yV 

5.  From  g^  take  |c.  Ans.  49c.  l^m. 
G.  Take  3ic.  irom  }  of2i$.  Ans,  43ic. 

7.  From  f  of  f  of  g5,  take  -f  of  96c.  added  to  ^  of  l^g. 

.^«s.  96c.  9*-ra. 

8.  From  intake  y\s.  j3nj.  49.  lid-. 

9.  From  -foz.  take  |pvvt.  *5ns.  13pwt.  12fa^r. 

10.  From  i  of  a  league  take  |  of  a  mile.  .'Z/is.  Imi.  Ifur 

11.  From  5  weeks  take  19j  days.  ^'^w^.  15da.  4ho.  48min. 

MULTIPLICATION  OF  VULGAR  FRACTIONS. 

Rule.* 

Reduce  compound  fractions  to  simple  ones,  and  mixed  numberf 
to  improper  fractions  ;  then  the  prodnct  of  the  numerators  will  be 
the  numerator,  and  the  product  of  the  denominators,  the  denomi- 
nator of  the  product  required. 

Note.  Where  several  fractions  are  to  be  multiplied,  if  the  nu- 
merator of  one  fraction  be  equal  to  the  denominator  of  another, 
their  equal  numerators  and  deoominators  may  be  omitted. 

Examples. 
1.  What  is  the  continued  product  of  4',  4,  1  off,  and  6. 

1X7" 

4^=¥>  }  of  1=. =^,  and  6=f 

4X8 

-whole  number  and  the  fraction  thus  formed  will  be  the  remainder :  but,  when 
the  numerator  in  the  fubtrahend  is  greater  than  that  in  the  minuend,  fubtract 
the  numerator  in  the  fubtrahend  from  the  common  denominator,  adding  the 
numerator  in  the  minuend,  and  carrying  i  to  the  integer  of  the  ful)trahend. 

Hence,  a  fracftion  is  fubtradted  from  .1  wiiole  nurnher,  by  taking  the  numera-* 
tor  of  the  fraAion  from  its  denominator,  and  placing  the  remainder  over  the 
denominator,  then  taking  one  from  the  whole  number. 

Examples. 
From  12|  12f  12 

Take    7i  7?  ?. 


Rem.     5'  4  A  n 


*  Multiplication  of  a  fraction  implies  the  taking  of  some  part  or  parts  -c^ii" 
the  multiplicand, and  therefore  may  truly  be  expressed  by  a  compound  fraction. 
Thus  1  multiplied  by  :i  is  the  same  as  4  of  Ji  ;  aud  as  the  directions  of  the  rule 
agfee  with  the  method  already  given,  to  reduce  these  fractions  to  simple  ones, 
it  is  shown  to  be  ri?.ht. 


88  VULGAR  FKACTiONS. 

13xlX  ''X6 

Then  V^XiX-^'^Xf— =III  =  1U  ^^^e  Answer. 

3X5X32X1  - 
'    2.  Multiply  -^  by  2^.  Ans.  //,, 

,1.  Multiply  51  by  }.  Ans.  |. 

4.  Multiply  i  of  6  by  I  off.  Ans.  /^. 

5.  Multiply  -f  of  f  by  |  of  i  of  1  If  Ans.  -j^^.. 

6.  Multiply  9|,  i  of  f,  and  12^  continually  together. 

Ans.  24i|. 

7.  What  is  the  continual  product  of  f  of  |,  5|,  7  and  |  of  f  ? 

Ans.  4^V- 
n.  What  is  the  continual  product  of  7,  J,  f  of  |,  and  3i  ? 

Ans.  i;^ 

Snother  method  for  the  Muliiplicaiion  of  mixed  Cluantilies. 

CASE  I. 

To  multiply  a  whole  number  by  a  fraction^  or  a  fraciion  by  a 
-whole  number. 

Rule. 

Multiply  the  whole  number  by  the  numerator  of  the  fraction  and 
tJivide  the  product  by  the  denominator  :  But  if  the  numerator  be 
1,  difide  by  the  denominator  only. 

3.  4.  5.  G.  7. 

28  36  48  325  25-1 

1  2  :l  5  t 

'S  a  4:  »  12 


1. 

2. 

Mult.  8 

15 

Byi 

i 

Prod.  2  7}  9}        3)72        4)144     8)1625    12)1813 

Prod.  24  36  203}     151  ^v^ 

CASE  11. 

To  multiply  a  zvkole  number  by  a  luixed  07ie. 

Rule. 
Multiply  by  the  fraction  as  in  Case  1st ;   l!?en  multijjly  by  the 
whole  number,  and  add  the   two  products,  as  h  the  examples — or, 
to  multiply  a  mixed  number  by  a  whole  one,  clvange   the  place  ot^ 
the  factors,  and  proceed  as  the  rule  directs. — St.e  example C 
1.  2.  3.  4.         .       5.  6. 

'Mult.  15  35  68  42  12:)  J  ,t 

By     3X.  5^  7ji  9-}  '  G,^  2.'i 

7A.  ]1|  748  126  645  Mull.  24 

^^        ]2^'       ^^'tV         18         l;af    r>y  i^. 

?r5d.    52A         186|  476  378  1032   15)l68~ 


Prod.     538-j.\  396  11121         11/^ 

£4 


IVod.  35.^—1. 


VULGAR  FRACTIONS,  89 

CASE  in. 

To  multiply  a  mixed  number  by  a  mixed  member. 

Rule. 

Multi(jly  the  integral  part  of  the  mnltipircand  by  the  denomina- 
tor of  its  fractional  part,  and  add  thereto  its  numerator:  Then 
mnUiply  by  the  mixed  multiplier,  by  Case  2d,  and  divide  the  pro- 
ddct  by  the  denominator  of  the  fractional  part  of  the  multiplicand, 
as  in  the  following  example. 
Mtdt.  422  j  1st.  42f=213^ 

By    Qdr  ^  which  mult,  by  8| 

After  this  manner  may  feet  and 

3)426     inches  be  multiplied,  calling  1  inch 

j^  of  a  foot,  2  inches  ^,  3  inches  {, 

142  }►  4  inches  i,  5  inches  jV,  G  inches  -}, 

1704     7  inches  -jV,  8  inches  |,  9  inches  ^ 

10  inches  |,  1 1  inches  {^  of  a  foot. 

1846 

I'roduct  =3694  j 

DIVISION  OF  VULGAR  FRACTIONS. 

Rule.* 
Prepare  the  fiactions  as  before:  then,  invert  the  divisor  and 
proceed  exactly  as  in  Multiplication:    The  products  will   be  the 
quotient  required. 

Examples. 
1.   Divide  ^  of  17  by  |  off 
1X17 

A  of  17  =  i  of  V  = =  y  and  |  of  f  =  i|  =  i  ;  there- 

3x  1 
17x2 

¥  ^  ^H  ^^'^  quotient  required. 

Ans.  1^^. 

3.  Divide  12}  bv  i  of  7.  '  Ans.  5>j, 

4.  Divide  b\-  by'7f  Ans.-*,}, 

5.  Divide  f  by  9.  Ans.  -jV- 

6.  Divide  i  of  }  of  f  by  1  of  f.  Ans.  *. 

7.  Divide  7  by  f  Ao^.  18|. 

8.  Divide  4204}  by  J  of  112.  Ans.  42|»f. 

*  The  reason  of  the  rule  may  be  shewn  thus.    Svippose  it  were  required  to 
4  2  4  J       4  4  2       ; 

divide  -  by  7.      Nnw  -  —  2  is  manifestly  "of -<?t  ^    ■■•';   but  r=i-of  2; 

12  ,  4  , 

thei'efoTe,-  of  2,  or  -,  must  be  contain«d  7  times  a^  often  in—  a5  2  that  is 

4x7 

gy~  ~'  the  a-nswcrjWklch  i«  according  to  ^c  rrfle. 

M 


ore, 

y-^ 

-i==- 

3X1 

2. 

Divi 

lo  4  by 

3, 

yo  UECiMAL  FKACTiONS. 

DECIMAL  FRACTIONS. 

A  DECIMAL  FRACTION  is  a  fraction  whose  denonilnator*  i? 
a  unit  wiih  &o  many  c\phers  annexed  as  the  numerator  has  places 
of  figure?:. 

As*  t'le  denomiuatcr  of  a  decimal  fraction  is  a]«a3'S  10,  or  100. 
or  1000.  &c.  Hie  denonnnaljrs  need  not  he  expre.^jsed  :  For  th.c 
numerator  only  mav  he  made  to  express  the  true  value  :  For  this 
purpose  it  is  only  required  to  uriie  the  numerator  with  a  point 
before  it  at  the  left  hand,  to  distinguish  it  from  a  whole  number, 
i^'hen  it  consists  of  so  many  ligures  a^-  the  denominator  has  cy- 
J^hers  annexed  to  uniiy,  or  1  :  So  -,-\  is  written  -5;  y%-^  -33  ;  jY/f 
'735,  iLc. 

The  point  prefixed  h  called  the  Seporafrix. 

But  ii  the  riumeratcr  has  not  so  many  places  as  tlie  denominator 
has  cyphers,  put  so  matiy  cyphers  before  it,  viz,  at  the  left  hand, 
as- will  make  up  the  defect ;  so  write  j|-^  thus,  05  ;  and  j-if%o  thus, 
•006,  &c.  Thus  do  tliese  fractions  receive  the  form  of  whole 
numhers. 

We  may  consider  unity  as  a  fixed  point,  from  whence  whole  num- 
bers proceed  infmitely  increasing  toward  the  left  hand,  and  deci- 
mals ifilinitely  decreasing  toward  the  right  hand  to  0,  as  in  thf 
following 

Table.^ 

i£  JC  J«  w:    c/j  . 


O 
9 

8 

-O  -3~ 

c    c 

re    cr 

a.     c/; 

■r    "Z    ~ 

c  ,o    c 

c  \sz  -a 

E  H  H 
7  6   5 

C 
C 

4 

0,' 

c 
3 

'^■Sg|||.l.2 

r-     ■>-     'I-     '^     ^     Z  ZZ  -Zl 
Q^     C     CU     ~   .C                   •  — 

2    1  -2  3  4  5  o  7  8   0 

*  It  will  be  x'cry  appnrcnt  to  tlie  learner  from  tlie  mture  of  decimals,  and 
what  has  been  said  of  Federal  il<fo«fj,  that  this  money  is  purely  decimal  ;  and, 
the  dollar  being  the  money  unit,  the  lower  denominations  are  plainly  so  many 
decimal  parts  of  a  dollar  ;  tlius  9  dollirs  and  8  dimes  are  expressed  9-8  .9  A 
doll. — 12  dollars,  4  dimes,  ui)d  7  cents  thus,  i2-47r=rr2  y  doll. — 20  dollars,  3 
dimes,  4  cents  and  5  mills,  thus  20345-20  3JL5.doll. — ico  dollars  and  9  mills, 
thus  100009  'OO,  _'>^  doll,  and  50  dollars,  5  cents,  thus  5005  — 50  .5  _  doll, 
wherefore,  it  is,  in  all  respfctK,  addtd,  subtracted,  multiplied  and  divided,  the 
same  as  decimals  ;  and,  of  all  coins,  it  is  the  most  simple.  • 

It  may  also  be  obst  rvcd  that  the  sum  exhibits  the  particular  number  of  each 
different  piece  of  money  contained  in  it,  viz.  455997  mills  -45599  "   cents  =• 

E.  D.  d.  cm. 
4559.^^"    dimes  —  455^9_s  t  dolhrs  —  45^fi_p.i)_^^^^  J  5  9  9   7- 

AKo,the  names  of  tl;c  coins,  less  llian  a  dollar,  are  significantpf  tluir  value? 
For  the  mUl,  winch  strands  in  tl.e  3d  place  at  the  rio;|.t  hand  of  the  separatxir 


DECIMAL  FRACnONS.  91 

From  this  table  it  ia  evident,  that  in  decimal*,  as  vvell  as  in  whole 
nunibers,  pach  figure,  takes  its  value  by  its  distance  trom  unit's 
'][>lace  :  If  it  be  in  the  first  place  alter  units  "l^or  the  separating 
point)  it  signifies  tenths  ;  if  in  the  second,  hundredths,  kc.  decreas- 
ing in  each  place  in  a  teiifuld  proportion. 

Every  single  figure  expressing  a  decimal,  has  for  its  denomina- 
tor a  unit  or  1,  with  so  many  cyphers  as  its  place  is  distant  irom 
unit's  place  :  Thus  2  in  the  decimal  part  of  the  table  =  /^  ;  3  = 
T'o 0  I  ^"^  ^^  Tuoo-' ^^'  ^^^  '^ '^  decimal  be  expressed  by  several 
rigures,  the  denominator  is  1,  with  so  many  cyphers  as  the  lowest 
iigure  is  distant  from  unit's  place.     So  -357  dignities  jV/u?  and  0053 

Cyphers,  placed  at  the  right  hand  ol  a  decimal  faction,  do  not 
alter  its  value,  since  every  signiticant  Iigure  continues  to  possess 
the  same  place:  So  -5,  50,  and  500,  are  all  of  the  same  vajue, 
anu  eacu  —  lo^— loo — Tooo  —  f 

But  cyphers,  placed  at  the  left  hand  of  a  decimal,  do  alter  its 
value,  every  cypher  depressing  it  to  ^^  of  the  value  it  had  before, 
by  removing  every  signiticant  figure  one  place  further  from  the 
place  of  uniis.  So  5,  -05,  005,  all  express  dilfercnt  decimals,  viz. 
•5,  t'o-  ;   '^5,  -jI^  ;  -005,  j^%^. 

IJence  may  be  observed  the  contrary  eftects  of  cyphers  being 
annexed  to  whole  numbers,  and  decimals. 

It  is  likewise  evident  from  the  table,  that  since  the  places  of  de- 
cimals decrease  in  a  tenfold  proportion  from  units  downwards,  so 
they  consequently  increase  in  a  tenfold  proportioo  from  the  right 
hand  toward  the  left,  as  the  places  of  whole  numbers  do  :  For,  tea 
hundredth  parts  make  one  tenth,  ten  tenths  make  1  ;  ten  units, 
ten;  ten  tens,  one  hundred,  &c.  viz.  yVd^^rV'  "io^^^»  ^^*^  1X10=^ 
10,  which  proves  that  decimals  are  subject  to  the  same  law  of  No- 
tation, and  consequently  of  operation,  as  whole  numbers  are. 

Decimal  fractions  of  unequal  denominators  are  reduced  to  one 
common  denominator,  when  there  are  annexed  to  the  right  hand 
of  those,  which  have  fewer  places,  so  many  cypi)ers,a8  make  them 
equal  in  phces  with  that  which  has  the  most.  So  these  decima!«^, 
•5,  06,  '455,  may  he  reduced  to  the  decimals,  "500,  '060,  and  '465, 
which  have,  all,  1000  for  their  denominator. 

Of  Decimals,  that  is  the  greatest,  whose  highest  figure  i^  great- 
est, wliether  they  consist  of  an  equal  or  unequal  number  of  places  : 
Thus,  -5  is  greater  than,  -459,  for  if  it  be  reduced  to  the  SJkipe  do- 
nominator  with  -459,  it  will  be   500. 


or  place  of  thousandths,  is  contracted  f com  milie  the  Latin  (or  thousand :  Ce/fi, 
which  occupico  tlie  second  place,  or  pUce  of  hundredths,  is  an  abbreviiticn 
of  cenlam,  the  Latin  for  hundred :  And  i//>w,  which  is  in  the  first  place  or  place  uf 
tenths,  is  derived  {i\ny\  d'ume,  ilie  French  for  tinth. 

Such  being  the  nafure  of  Federal  Money,  its  operations  can  in  no  ot'icr  way  Ire 
;.o  well  understood  as  in  obtaining  a  good  knowledge  of  decimals,  and  applyiii^j 
rhcir  several  rules  to  the  various  cases  of  monev  matters. 

In  sums  of  Federal  Money,  it  is  customary  to  read  it  in  dollars,  cents  and 
mills.     Thus  the  ubove  example  is  read  /J55  dolls.  99  cents  and  7  mills. 


9^  DECIMAL  FRACTIONS. 

A  mixed  number,  viz.  a  whole  number,  with  a  decimal  annexed, 
is  equal  to  an  improper  fraction,  whose  numerator  is  all  the  figuress 
of  the  mixed  number,  taken  as  one  whole  number,  and  the  denom- 
inator, that  of  the  decimal  part.  So  45*309  is  equal  to  VoW,  as 
is  evident  from  the  method  given  to  reduce  a  mixed  number  to  an 
improper  fracliop : 

Thus,  4oxlOOO-'r-309=''/(?/o^  ^^  above. 

ADDITION  OF  DECIMALS, 

1.  Place  the  numbers,  whetlier  mixed,  or  pure  decimals,  under 
each  other,  according  to  the  value  of  their  places. 

2.  Find  their  sum  as  in  whole  numbers,  and  point  oft"  so  many 
places  for  decimals,  as  are  equal  to  the  greatest  number  of  deci- 
mal places  in  any  of  the  given  numbers. 

Examples. 

1.  Find  the  sum  of  19  073-l-2-3597+223+-0197681-|-347eiT- 
12-338. 

19073 
2-3597 
223- 

•0197581 
3478  1 
12-358 


3734.9104581  the  sum. 


2.  Required  the  sum  of 429-4-21  •37-f-355003^}-l  07+1  7  ?' 

Ans.  808•148 
3.  Required  the  sum  of  5-3-|- 1 1  •973-f49H-94- 1  •7314H-343  f 

Ans.  103-2044^ 
.   4.  Required  the  sum  of  973-|-19-|-l-75-f  93-7164-f-9501  ? 

Ans.  1088.'U65, 

SUBTRACTION  OF  DECIMALS. 
Rule. 
riacp  the  numbers  according  to  their  value  ;  then  subtract  as  in 
whole  numbers,  and  point  oil  the  decimals  as  in  Addition. 

Examples. 

I.  Find  the  difference  of  1793-13  and  817C5G93^ 
From  1793-13 
Take    817  05C93 


P.emainder     976*07307 

i.  From  17  J -195  take  125-917C.  Ans.  45  2  774. 

75.  From  219-1384  take  195-01.  Ans.  23-2284. 

a.  From  480  take  21 3-0075.  Ans.  234-9f>2,5. 


DECIMAL  FRACTIONS,  93 

MULTimCATION  OF  DECIMALS. 
I  CASE  I, 

Rule. 

1.  Whether  they  b^  mixed  numbers,  or  pure  decimals,  place  the 
factors  L'.nd  multiply  tliem  as  in  whole  numbers. 

'2.  Point  off  so  many  figures  from  the  product  as  there  are  deci- 
mal places  in  both  thi*  factors  ;  and  if  there  be  not  go  many  places 
ill  the  product,  supply  the  defect  by  prefixing  cyphers. 

The  reason  for  pointing  off  the  figures  for  decimals,  is  evident 
from  substituting  the  equivalent  vulgar  fractions,  as  in  the  ibllow- 
ing  example-  Or,  'Hie  reason  of  the  rule  for  pointing  off  the  fig- 
ures for  decimals,  is  evident  from  the  notation  of  decimals.  Thus 
.5X-5=-25  ;  for  -Sxl^S,  or  once  five  tenths.  But,  as  the  multi- 
plier is  less  than  unity,  or  tenths,  multiplying  is  only  taking  tenths 
o^  tenths,  and  so  ma^iy  tenths  of  tenths  are  evidently  so  niany  hun- 
dredths. So  also,  tenths  ofhundredths  would  be  thousandths  ;  hun- 
dredths of  hundredths,  would  be  ten  thousandths,  and  so  on. 

Examples. 
1.  Multiply   02345                                             23^5 
by  -00163  -02345= 


100000 


7035                  163 
14070  .00163= 


2345  lOOOOe 


382235 
•0000382235  the  product,: 


10000000000 


S.  Multiply  25-238  by  12-17.  Ans.  307-14646. 

'6.  Multiply  -3759  by  -945.  Ans.  -3552255, 

4.  Multiply  -84179  by  -0385.  Ans.  -032408915. 

To  multiply  by  10,  100,  1000,  &c.  remove  the  separating  point 
.350  many  places  to  the  right  hand,  as  the  multiplier  has  cyphers- 

(10      )  (3  45 

So  -345  Multiplied  by  \  100    \  makes  \  345 
(1000)  (345  ' 

For  -345x10  is  3-45P,  &c. 


CASE  II. 

To  contract  the  operalion^  so  as  to  retain  so  many  decimal  places  \u 
the  Product  ««  moy  be  thought  necessary. 

Rule. 

1.  Write  the  unit's  place  of  the  multiplier  under  that  figure  of 
the  multiplicand,  whose  place  you  would  reserve  in  the  product ; 
and  dispone  of  the  rer^t  of  the  figures  in  a  contrary  order  to  what 
they  are  usually  placed  in. 

2-  In  multiplying,  reject  all  the  figures  which  are  on  the  right 
hand  of  the  multiplying  digit^  and  ict  down  the  products,  po  tha^ 


m 


DECIMAL  FKACTiONiS. 


their  right  hand  figures  may  fall  in  a  straighlt  line  below  each  ether  ; 
observing?  to  increase  the  first  figure  of  evesrj  line  with  what  would 
arise  b}^  carrying  1  from  5  to  15,  2  from  I,':"  to  26,  3  from  25  to  35, 
kc.  from  the  preceding  figures,  when  you  begin  to  multiply,  and 
the  sum  will  be  the  product  required. 

Examples. 

1.  It  is  required  to  multiply  5G-7534916  %  6-376928,  and  to  re- 
ia'm  only  live  place©  of  decimals  in  the  product. 

Contracted  way.  Common  way. 

66-7534916  66-75340 16 


129673 


6  376928 


28376746  . 
1702605  . 

397274  . 

.34052  . 

5108   . 

113  . 

45  . 


305- 15943 


45 

40279328 

113 

5069832 

6107|314244 

34052  09496 

397274  4412 

1702604  748 

8376745  80 

305-15943  80818048 


By  the  operation  in  the  common  umy,  it  is  evident  that  all  the 
ngures  which  are  cut  off  at  the  right  hand,  by  the  perpendicular 
iine,  are,  wholly  omitted  in  the  coniracted  way,  and  the  last  product 
here  is  the  first  there  ;  consequently  the  rea^^on  of  placing  the 
?nultip!ier  in  a  reverse  order,  must  appear  very  plain. 

difisiOj^  of  decimals. 

Rule.* 

,  I,  The  places  of  decimal  parts  in  the  divisor  and  quotient  cu-d'..i- 
ed  togeth.er,  must  always  be  equal  to  those  in  the  dividend  ;  there- 
fore, divide  as  in  whole  numbers,  and,  from  the  right  hand  of  the 
qtiotient,  point  ofi' so  many  {daces  for  decimals,  as  the  decimal  pla- 
ces in  the  dividend  exceed  those  in  the  divisor. 

2.  If  the  places  of  the  quotient  be  not  so  many  as  the  rule  re- 
quires, stippiy  the  defect  liy  pretixing  cyjjhers  to  the  left  hand. 

3.  If  at  any  time  there  be  a  remainder,  or  the,  tlecimal  places  in 
tiie  <iivi.*r  be  more  tfian  those  ia  the  diviilend,  cyphers  may  be 
annexed  to  the  dividerid,  or  to  t*!ie  remainder,  and  the  q-juiient 
.  irriCu  on  to  ativ  ile^jce  of  exactness. 


*  Tbe  reason  of  pointing  off  »o  many  decimal  places  in  the  quotient,  as 
Cixohc  in  the  dividend  exceed  tliDsc  in  the  divisor,  will  easily  ;ippear,  for,  s.ince 
iiic  number  of  decimal  places  in  the  dividend  is  cquHl  to  those  in  the  divisor 
and  quotient  taktn  together,  hy  the  nature  of  nmhipliration  :  It  therefore  fc!- 
,  vw.;  rhnf  thf'rv^'-  "'  '-^v-rainj  SO  miuy  as  the  tlvid'.nd  tx^ceeds  the  divisor. 


DECIMAL  FRACTiOXS.  95 

Examples. 
1.  2. 

219)-1 17841O75(-0OO538O87,  Lc.  •3719)38-0000(102-,173,  kc 

1095  In  Example  1st,  the  divi-  3719  ^V 

sor  having  no  decimals,  the  

834  quotient  must  have  so  ma-       8100 

G57  ny  as  there  are  in  the  divi-       7438 

dend.     In  Example  2,  tl«e       

1771  dividend    being  an  integer         Q.G^O 

1752  must  have  at  least  so  many  3719 

cyphers  annexed    as  there  ■ 

1907     are  decimals  in  the  divisor,         29010 
1752     and  so  tar  the  quotient  will         26033 

he  whole  numbers,  then  an- 

1555  nexing  more  cyjdiers,  the 
1533  remaining    lignres    in    the 

quotient  will  be  decimal:*, 

22  according  to  the  Rule.  18 

3d.   133)5737(43-1353-{-  (4th.)     £3'7)G5321(275G-16-i 

5th.  172)918  217(12753-1-  (6th.)  25-17)315-G293(1253-{- 

7ih.  •317)29-417(92-f-  (oth.)  37-9)-0059374(     loG-j- 

9th.   •375)-173948375(4G3862-f 

Having  a  multiplier,  to  find  a  divisor -Si-hicli  shall  give  a  quotient  equal 
to  the  product  by  that  multiplier. 

Rule. 
Divide  unity  by  the   given  multiplier,  and  the  quotient  will  he 
the  divisor  sought. 

What  divisor  is  that,  by  which  dividing  5357,  shall  give  a  quo- 
tient equal  to  the  product  of  the  same  number  multiplied  by  250  ? 
250M-000(-004  the  Answer.     And  -004)5357  000(1339250- 

Proof.     6357x250=1339250- 

Having  a  divisor,  to  find  a  midiiplier  ii-hich  shall  give  «  product  equal 
to  the  quotient  by  that  divisor. 

Rule. 
Divi<le  unity  b}^  the  given  divisor,  and  the  quotient  will  be  the 
mulli[)lier  sought. 

What  multiplier  is  that,  by  which  multiplying  5357,  shall  give  a 
product  equal  to  the  quotient  of  the  same  number  divided  by  004 

Ans.*250. 

CASE  II. 

To  contract  Division,  tuhen  there  are  many  decimals  in  the  dividend. 

and  the  divisor  is  large. 

Rule. 
1.  Whatever  place  of  the  dividend  corresponds  with  the  unit  £ 

f  The  following  ovjesllons  arc  left  unpointed  hi  the  ciiotieni:  iltj  etercise  t?  r 
Iieamer, 


^^  DECIMAL  FRACriONS. 

place  of  the  divisor,  at  tlie  first  step  of  ihe  division,  the  same  place 
must  the  first  figure  of  the  quotient  have. 

2.  Ill  dividing  reject  the  last  right  hand  figure  of  the  divisor,  at 
erery  step,  (instead  of  bringing  down  a  figure,  as  i«  common.)  and 
make  the  last  remainder  the  dividend  for  the  new  divisor  at  every 
step:  Thus  continue  the  division  until  the  divisor  shall  be  ex- 
hausted. 


99-5678)4-6789837368(  0409931  Quotient. 
3-982712 


!>9'567)G96271  Here,  the  units  place  of  the  divisor 

597402  in  the  first  step  falls  under  7  in  the  place 

of  hundredths  in  the  dividend,  (here- 

^t9'56)  98869         .  ,    fore,  I  put  4,  the  first  quotient  figure, 

89604  in  the  place  of  hundredths,  by  prefixing 

a  cypher. 

:^9'5)  9265  1  have  set  down  every  divisor,  to  ex* 

8935  plain  the  work  ;  but  you  need  only  put 

a  dash  over  every  figure  rejected,  a^ 

0.9)  310  you  pfoceed,  to  show  it  is  omitted. 
297 

9)   13  '' 

9 

Remainder  4 
When  decimals  or  whole  num- 
bers are  to  be  divided  by  10,  100,  Exaiui 
1000,  &c.  (viz.  unity  with  cyphers) 
it  is  performed  by  removing  the 
9eparatrijf,inthe  dividend,  so  ma- 
ny places  toward  the  left  hand  as 
there  are  cyphers  in  the  divisor. 

REDUCTION  OF  DECIMALS. 

CASE  I. 

To  reduce  a  Uulgar  ^(action  to  its  equiva'^'nt  Decimal. 

Rule.* 

DiviJe  the  numerator  by  the  denominator,  as  In  division  of  do 

»imal*,  and  the  quotient  will  be  the  decimal  required  : — Or,  su 

many  cyphers  as  you  annex  to  the  given  numerator,  so  many  pla- 

*  By  annexing  one,  two,  three,  &c.  cyphers  to  the  numerator,,  the  value  of  the 
traaion  is  increafcd  ten,  a  hundred,  &c.  times.  After  dividing,  the  quotient 
will  of  courfe,  be  ten.a  hundred,  &o.  times  too  much  -.  the  quotient  muft  therc- 
*bre  be  divided  by  ten,  a  hundred,  &c.  to  give  the  true  quotient  or  fraaioo.  In 
the  f:rft  example,  ^,  is  mfldc  i  «LP  «  arr » 3.*  ,  which  divided  bv  lOOU,  is  _.yj,T. 
%''i25;  and  is  t1»c  ntlc. 


0-. 


DECIMAL  FRACTIONS,  97 

ces  must  be  pointed  off  in  the  quotient,  and  if  there  be  not  so  many 
places  of  figures  in  the  q'lotient,  the  deficiency  must  be  supplied 
by  prefixing  so  many  ciphers  before  the  quotient  figures. 

Examples. 

1.  Reduce  i  to  a  decimal.  8)1.000 

•125  Ans. 

2.  Reduce  f ,  f  and  |  to  decimals.     Answers,  STS^  625,  -6664- « 

3.  Reduce  ^,  i,  |,  i,  |,  f  and  |  to  decimals. 

Answers,  -25,  -6,  75,  -SSS-f ,  -8,  -833+,  -875. 
I.  Reduce  -,%,  f|,  4W,  and  /g-  to  decimals. 

Answers,  -263+,  •692-f ,  -025,  -25. 
5.  Reduce  ^^j,  j-f^^,  ahd  ^/tt  to  decimals. 

Answers,   0186+,  -00797+,  -00266+. 

CASE  H. 

To  reduce  numbers  of  different  denominations^  as  of  Money^  Weight 
and  Measure^  to  their  equivalent  decimal  values. 

Rule.* 

I.  Write  the  given  numbers  perpendicularly  nnder  each  other, 
for  dividends  ;  proceeding  orderly  from  the  least  to  the  greatest. 

II.  Opposite  to  each  dividend,  on  the  left  hand,  place  such  a 
number,  for  a  divisor,  as  will  bring  it  to  the  next  superiour  denom- 
ination, and  draw  a  line  perpendicularly  between  them. 

III.  Begin  with  the  highest,  and  write  the  quotient  of  each  di- 
vision as  decimal  parts  on  the  right  hand  of  the  dividend  next  be- 
low it,  and  so  on,  until  they  are  all  used,  and  the  last  quotient  will 
be  the  decimal  sought. 

Examples. 
1.  Reduce  17s.  8fd.  to  the  decimal  of  a  pound. 


4 
12 
20 


3- 


8-75 


17-729166,  &c, 

•886458,  &c.  the  decimal  required* 


Here,  in  dividing  3  by  4,  I  suppose  2  cyphers  to  be  annexed  to 
the  3,  which  make  it  300,  and  -75  is  the  quotient,  which  I  write 
against  8  in  the  next  line  ;  this  quotient,  viz.  875  feing  pence  and 
decimal  parts  of  a  penny,  I  divide  them  by  12,  which  brings  them 
to  snillings  and  decimal  parts,  1  therefore  divide  by  20,  and,  there 
being  no  whole  number,  the  quotient  is  decimal  parts  of  a  pound. 

*  The  reason  of  the  rule  may  be  explained  from  the  first  clample ;  Thus, 
three  farthings  are  ^  of  a  penny,  which,  reduced  to  a  decimal,  is, -75;  conse- 
quently, 8fd.  may  be  expressed,  8. 75d.  but  8.75  is  81|  of  a  penny  =  y3_7_5_  of 
a  shilling,  which  reduced  to  a  decimal,  is,  •7291663.+  In  like  manner, 
17749166$.+  are  i  i/J?^J^^6,.  ^  ii^|a  ^^^j^  ==  .88/>4^8:+-a»b7  the  role. 

w 


98  DECIMAL  FRACTIONS. 

2.  Reduce  1,  2,  3,  4,  and  so  on  to  19  shillings,  to  decimals. 
Shillings.     123450789  10 
Answers. -05,       -1,     -16,       -2,     -25,       -3,     -36,      -4,    -45,  '5, 
Shillings.      11         12        13         14        15        16         17        18  19 
Answers.    -55,        -6,      -65,       -7,      -75,        -8,      -85,       -9,  -95. 

Here,  when  the  shillings  are  even,  half  the  number,  with  a  point 
prefixed,  is  their  decimal  expression  ;  but  if  the  number  be  odd, 
annex  a  cypher  to  the  shillings,  and  then  halving  them,  you  will 
have  their  decimal  expression. 

3.  *Keduce  1,  2,  3,  and  so  on  to  11  pence,  to  the  decimals  of  a 
shilling. 

Pence.  12  3  4  5  6 

Answers.       -083-}-,     -166,  -25^       -SSS-f,     -416-1-,         -5, 

Pence.  7  8  9  10  11 

Answers.        •583-f,         -Oee-f,         -75,  -8334-,         -916+. 

4.  Reduce  1,  2,  3,  &c.  to  11  pence,  to  the  decimals  of  a  pound. 
Pence.  1  2  3  4  5 
Answers.      -00416+,     -00834-,     '0125,       •Oi666-f,     -0208+, 
Pence.             6              7              8              9              10             11 
Answers.     -025,  -02916+,  -0333+,   -0375,    0416+,  -04583+. 

5.  Reduce  1,  2  and  3  farthings  to  the  decimals  of  a  penny. 

lqr.  =  -25d.  2qr.  =  -5d.  and  3qr  =-75d.  Answers. 

6.  Reduce  1,  2  and  3  farthings  to  the  decimals  of  a  shilling. 
Answers,  lqr.^-02083+s.  2qrs.  =  -04166+s.  3qrs.=-0625s. 

7.  Reduce  1,  2  and  3  farthmgs  to  the  decimals  of  a  pound. 
Ans.  lqr.  =  -0010416+£.  2qra. =-002083+ £.  3qrs.=-003125£. 

8.  Reduce  13s.  5Jd.  to  the  decimal  of  a  pound      Ans.  -6729+. 

9.  Reduce  7Cwt.  3qrs.  17ib.  lOoz.  12dr.  to  the  decimal  of  a  ton. 

Am.  -39538+. 

10.  Reduce  lOoz.  13pwt.  9gr.  to  the  decimal  of  a  pound  Trow 

Ans.  -8890625." 

11.  Reduce  3qrs.  3n.  to  the  decimal  of  a  yard.     Ans.  '9375. 

12.  Reduce  5fur.  12po.  to  the  decinial  of  a  mile.    Ans.  '6625^ 
33.  Reduce  55m.  37sec.  to  the  decimal  of  a  day. 

Ans.  -03862+ . 

CASE  III. 

To  find  the  decimal  rf  any  number  of  shillings,  pence  and  farthings, 
by  inspection. 

RuLE.t 
1.  Write  half  the  greatest  even  number  of  shillings  for  the  first 
decimal  figure. 

*  The  answers  to  this  question  arc  the  same  as  the  decimal  parts  of  a  foot. 

•j-The  invention  of  the  rule  is  as  follows  :  As  shillings  are  30  many  loths  of 
a  pound,  half  of  them  must  be  so  many  tenth's,  and  consequently  take  the 
place  of  tenths  in  the  decimals;  but  when  they  arc  odd,  their  half  will  always 
consist  of  two  figures,  the  first  of  which  will  be  half  the  even  number,  next 
less,  and  the  second  a  5  ;  Again,  farthings  are  so  many  96oths  of  a  pound, and 
had  it  happened  that  icoo,  instead  ©f  960,  had  made  a  pound,  it  is  plain  any 


.     DECIMAL  FRACTIOKS.  <)9 

II,  Let  the  farthings  in  the  given  pence  and  farthings  possess  the 
secpnd  and  third  places  :  observing  to  increase  the  second  place 
or  place  of  hundredths,  by  5  if  the  shillings  be  odd,  and  the  third 
place  by  1,  when  the  farthings  exceed  12,  and  by  2  when  they  es- 
ceed  36. 

Examples. 

1.  Find  the  decinaal  of  13.s.  9|d.  by  inspection. 

•6..  =1  of  12s. 
6    for  the  odd  shilling. 
39  =the  farthings  in  93d. 
Add  2  for  the  excess  of  3C. 

•691  =  decimal  required. 

2.  Find,  by  inspection,  the  decimal  expressions  of  18s.  3id.  and 
17s.  8id.  Ans.  £-914  and  £-885. 

4.  Value  the  following  sums,  by  inspection,  and  find  their  total, 
viz.  15s.  3d,4-8s.  lUd.-f  10s.  G^d.-f-ls.  S-^d.+^d.+^^d. 

Ans.  "je  1-834  the  total. 

CASE.  IV. 

To  find  the  value  of  any  given  decimal  in  the  terms  of  the  integer. 

Rule. 

L  Multiply  the  decimal  by  the  number  of  parts  in  the  next  less 
denomination,  and  cut  off  so  many  places  fof  a  remainder,  to  the 
right  hand,  as  there  are  places  in  the  given  decimal. 

H.  Multiply  the  remainder  by  the  nei^t  inferiour  denomination^ 
and  cut  off  a  remainder  as  before. 

III.  Proceed  in  this  manner  through  all  the  parts  of  the  integer, 
and  the  several  denominations,  standing  on  the  left  hand,  make  the 
answer.  This  case  is  the  reverse  of  Case  U.  and  the  reason  of 
the  rule  is  hence  obvious. 

Examples. 
1.  Find  the  value  of  '73968  of  a  pound.  , 

20 


14-79360 
12 

9*52320 
4 


2»0928O  Ans.  14s.  9id. 

number  of  farthings  would  have  made  so  many  thousandths,  and  might  have 
taken  their  place  in  the  decimal  accordingly.  But  960  increased  by  J-  part, 
of  itself,  is :=:iooo,  consequently  any  number  of  farthings,  increased  by  their 
_i_  part,  will  be  an  exact  decimal  expression  for  them  :  Whence, if  the  number 
of  farthings  be  more  than  la,  oj  P*""'  '^  greater  than  jqr.  and,  therefore  i  must 
be  added ;  and  when  the  number  of  farthings  is  more  than  z^t  -^^  P*rt  U 
greater  than  i^qf.  for  which  2  must  be  added. 


100  DECIMAL  FRACTIONS. 

2.  What  is  the  value  of -679  of  a  shilling?  Ans.  8-148d. 

3.  What  is  the  value  of -DgGDje?         Ans   19s.  llfd/i^  or£l. 

4.  Wiiat  is  the  value  of -617  of  a  Cvvt.  ? 

Ans.  2qrs.  131b.  loz.  lO^^dr. 

5.  What  is  the  value  of  -8593  of  a  lb.  Troy  ? 

Ans.  lOoz.  6pwt.  5gr. 

6.  What  IS  the  value  of -397  of  a  yard?        Ans.  Iqr.  2-352n. 

7.  What  is  the  value  of  '8469  of  a  degree  ? 

Ans.  5Sm.  6fur.  35po.  Oft.  llin. 

8.  What  is  the  value  of '569  of  a  year  ? 

Ans.  207da.  16ho.  26m.  24sec. 

9.  What  is  the  value  of  -713  of  a  day  ?     Ans.  17h.  6no.  43sec. 

CASE  V. 

To  find  the  value  of  any  decimal  of  a  pound  by  inspection. 

Rule.* 

Double  the  first  figure,  ov  place  of  tenths,  for  shillings,  and  if 
the  second  figure  he  5,  or  more  than  5,  reckon  another  shilling  ; 
then,  after  the  3  is  deducted,  call  the  figures  in  the  second  and 
third  places  so  many  farthings,  abating  1  when  they  are  above  12, 
and  2  when  above  36,  and  the  result  will  be  the  answer. 

JVote.  When  the  Decimal  has  but  2  figures,  if  any  thing  remain 
after  the  shillings  are  taken  out,  a  cypher  must  be  annexed  to  the 
right  hand,  or  supposed  to  be  so. 

Examples. 

1.  Find  the  value  of  •876£  by  inspection. 
16s.       =  double  of  8. 
Is.       for  the  6  in  the  second  place,  which  is  to  be  taken 
And  6id.=26  farthings  remain  to  be  added.  [out  of  7. 

Deduct  ^d.  for  th^  excess  of  12. 


17s.  6id.  the  Ans. 


2.  Find,  by  inspection,  the  value  of  •49J6. 
8s.  -  -  =  double  of  4. 
Is.  -  -  for  the  5  in  the  place  of  hundredths. 

lOd.  =  40  farthings,  a  0  being  annexed  to  the  remaining  4 
Ded.         id.  for  the  excess  of  36. 


9s.  9id.  the  Answer. 

3.  Find^the  value  of  •097je  by  Inspection.  Ans.  U    n]d. 

4.  Value  the  following  decimals  bv  Inspection,  and  find  their 
?nm,  viz.  •785£  -f  •537i:  -f  '916£  +  •74i:  +  '5£  +  'SojC  -f- 
•09JS  4-  -OOSje.  Ans.  £3  16s.  6d. 


•  As  this  rule  is  the  reverse  of  the  rule,  Case  HI.  the  reason  is  apparent  from 
the  demonstration  of  that  rule. 


DECIMAL  FRACTIONS. 


201 


i DECIMAL  TABLES  OF  COIN.  WEIGHT.  ANDMEASURE. 


TABLE  I. 

Coin. 

£  1  the  Integer. 


dec. 
•95 


Shil 

19 

i8 

17 

i6 

^5    '75 

Ul  7 
'6s 
•6 
■55 
•5 


Shil 

9 

8 

7 
6 

5 
4 

3 

2 

X 


dec. 
•45 
•4 
•35 
•3 
*»5 
•a 

•15 
•I 

•05 


TABLE  III. 

Troy  Weight. 

lib.  the  Integer. 

Ounces  the  fame  as 

Table  II 


Pence. 
It 

10 

9 
S 

7 
6 

5 
4 
3 

a 

1 


Pwts.  Decimals. 

10  '041666 

9  -0375 

8  -035333 

7  '029166 

6  -025 

5  -020833 

4  -016666 

3  •01*5 

a  -008333 

Decimals.         1  '004166 

•045833 ioTalns.  DecTniaJsr 


•C41666 
•0375 
•033383 
•019166 

•025 

'020833 

•016666 

•0125 

•008333 

•004166 


Earth's 

3 

2 

I 


Decimals. 
'003125 
'0020833 
•0010416 


12 
II 

10 

9 
8 

7 
6 

S 

4 
3 

2 

I 


'O02083 

•00191 

•001736 

•001562 

'001389 

'001215 

•001042 

•000868 

•000694 

•000521 

•000347 

•000173 


Pounds. 
27 
26 
a5 
24 
^3 
22 

21 

20 

19 
18 

17 
16 

Jt5 
14 
13 

12 

XX 

10 

9 
8 

7 
6 
5 
4 
3 
2 
t 


Decimals 
•241071 
•232143 
•223214 
•214286 
•ao5357 
•196428 
•1875 
'178571 
•16^643 
•160^14 
•151786 
•142857 
•1339*8 
•125 
•110671 

•107143 

•098x14 

'089286 

•080357 

'071428 

'0625 

•0535^1 

'044643 

•035714 

'026786 

•017857 

•008928 


Ounces.  I  Decimals. 
12       .-75 


XI 

10 

9 
8 

7 
6 

5 
4 
3 
2 

X 


•6875 

'625 

•5625 

•5 

•4375 

•375 

'3125 

•25 

•1875 

•125 

'0625 


j  Drams.  I  Decimals. 
15         -058593 


TABLE  II. 

Coin  &LongMeas. 

I  Shil  and  i  Fool 
the  Integer. 

P^°«^  Decimals. 


Inches. 

IX 


7 
6 

5 
4 
3 
2 
I 


•916666 

•833333 

•75 

•666666 

•583333 

•5 

'4166661 

•333333 

•15 

•166666 


I  Oz.  the  Inttgcr. 
Pennyweights  the 
fame  as  Shillings  in 
the  firft  Table. 

Decimals. 
•025 
•022916 
•020833 
•01875 
•016666 

•014583 

•0125 

•010416 

•008333 
•00625 
•004166 
•002083 


Grains. 
12 
XI 
10 

9 
8 

7 
6 

5 
4 
3 

2 

I 


Ounces.  jDecimals. 
15  -008370 
14  1  •007812 
13  ,  -007254 
X2  I  '006696 
11       I    -006138 


14 
13 
12 
II 

10 

9 
8 

7 
6 

5 
4 

3 
2 

I 


•054687 

•050781 

046875 

•642968 

•039062 

•035T56 

•03125 

027343 

•023437 

•019531 

•015625 

•011718 

•007812 

'003906 


10 

9 
8 

7 
6 

J 
4 
3 

2 

I 


1  '00558 
•005022 
•004464 
•003906 
'003348 
'00279 
•002232 
'001674 
•001116 
•000558 


TABLE  VI 

ClothMeasuure 
I  Yard  the  Integer 
Decimals. 

•75 

•5 

•*5 


Farth's 

3 

2 

X 


TABLE  IV. 
Avoirdupois  Wt. 
•0833331  iialb.  the  Integer. 


qr.ofoz8.|Dccimals 
3      '  '000418 

2  '000279 

T       I    •000139 


Decimals, 
•0625 
■041666 
•020383 


Qrs. 

3 

2 

I 


Decimals. 
'75 
•5 
•25 


Qrs. 

3 

2 

X 


Nails.  Decimals. 

3 

•1875 

2 

•125 

I           '0625 

TABLE  VII 
Long  Measure. 
I  Mile  the  Integer, 
Decimals. 
^•568182 
•511364 


Yard! 

1000 
900 


TABLE  V. 
AVOIRUUFOIS  Wt. 
lib.  the  Integer. 
Decimals. 


Ounces 
15 
14 
13 


•9375 

•875 

-8125 


800 
700 
600 
500 
4C0 
300 
200 
100 


•454545 

•397717 

•34 

•284091 

•227272 

•^70454 

•113636 

•056818 


Carried  over. 


102 


DECJIVIAL  FRACTIONS. 


Yards. 

becimals. 

TABLE  Vlir. 

Days. 

30 

20 

Decimals. 

1 

90 
80 

•(J  5 1136 
•045454 

LiquiD  Measvre. 

•082192 
054794 

Hours. 
11 

Decimals. 

458333 

70 

•039773 

I  Gal.  the  Integer. 

10 

•027397 

10 

•416666 

60 

034091 

Quarts  the  fame  as 

9 

•024657 

9 

•375 

50 

40 

028409 
•022727 

qrs.  in  Table  VI. 
■  Pint.      -1'^''^ 

8 

7 

021918 
019178 

8 

7 

•333333 
•291666 

30 
20 

•017045 
011364 

o  Gills. 

2 

•09375 
•0625 

6 
5 

•016438 
013698 

I 

•25 
•208333 

10 

•005682 

1 

nQirtr 

4 

•010959 

4 

•166666 

9 

.0051 14 

1                      yj^ic-y 

3 

•008219 
•005479 

3 

■125 

8 

•004545 

TABLE  iX. 

2 

2 

•083333 

7 

•003977 

1 

•0f)9739 

1 

•041666 

6 

3 

2 

•003409 
.002841 
.002273 
.001704 
.001136 

Time. 
1  Year  the  Integer. 
Months  the  fame  as 
Pence  in  Table  II. 

1 

I 

iDay 

Hours. 

23 

the  Integer. 

Decimals. 

•958333 

minutes 
30 
20 

.  Decimals. 

•020833 
•013888 

1 

000568 

Days. 

Decimals. 

22 

•916666 

10 

•006944 

Feet. 

2 
1 

Decimals. 

■0G03787 
0001892 

365 
300 
200 
100 
90 

roooooo 

•821928 
•547945 
•273973 
•246575 

21 
20 
19 
18 
17 

•875 

•833333 

•791^66 

•75 

•708333 

9 
8 
7 
6 
5 

•00625 

•005555 

•004861 

•004166 

003472 

laches. 

Decimals. 

6 

0000947 

80 

•219178 

16 

•666666 

4 

•002777 

5 

•000079 

70 

•191781 

15 

•625 

3 

•002083 

4 

•0000632 

.     60 

•164383 

14 

•583333 

2 

.001388 

3 

•0000474 

50 

•136986 

13 

•541666 

1 

•000694 

2 

0000316 

40 

109589 

12 

•5 

1 

•0000158 

1 

General  Rule. 


To  find  the  value  of  goods  in  Federal  Money. — Multiply  the 
price  and  quantity  together;  poi.jt  off  in  the  product,  for  denomi- 
nations lower  than  dollars,  as  nrjany  places  as  there  are  in  the  giv- 
en price  ;  or,  if  there  be  decimal  places  in  the  quantity  also,  ac- 
cording to  the  rule  for  multipllicatioi\  of  decimals.  This  is  reaUy 
mulliplication  of  decimals,  the  dollar  being  considered  the  unit. 

Examples. 

1.  Find  the  cost  of  823  yards,  at  .^1.29c.  a  yard. 

823xg  1.29c.  =^106 1.67c.  Ans. 

2.  Find  the  value  of  56yds.  2qrs.  at  ,^3.11r.  per  vard. 

56yds.  2qr8.=56-5;  and  66-5X3-1  l=^175.7"^Ic.  5m.  Ans. 
•;.  What  must  I  pay  for  6iyda.  at  ^5.50c.  per  yard  ' 

Ans.  ,$33.68c.  7ro.  -5. 
4.  Bought  7}}yd3.  at  34 cents  per  yard:  what  did  I  pay  for  the 
vffcole  ?  Ans.  g2.62J  |e. 


COMPOUND  MULTIPLICATION.  lOr* 


COMPOUND  MULTIPLICATION* 

IS  extremely  useful  in  finding  the  value  of  Goods,  &c.  And  as 
in  Compound  Addition,  we  carry  from  the  lowest  denomination  to 
the  next  higher,  so  we  begin  and  carry  in  Compound  Multiplication  : 
One  general  rule  being  to  multiply  the  price  by  the  quantity.  The 
reason  of  the  following  rules  is  obvious. 

CASE  I. 

fVhen  the  quantity  does  not  exceed  12  yards, -pounds yS,'c  :  Set  down 
the  price  of  1,  and  place  the  quantity  underneath  the  least  denomi- 
nation, for  the  multiplier,  and,  in  multiplying  by  it,  observe  the 
same  rules  for  carrying  from  one  denomination  to  another  as  ia' 
Compound  Addition. 

Introductory  Examples. 


Multiply 
by 

15 
31 

I. 

S. 

17 
14 

d. 
1 

2 

2 

£ 

25 

2. 

8. 

12 

d. 
8 
3 

D.  d. 

8     5 

3. 
c. 

1 

m. 

7 
4 

4. 

£     s.      d. 
67     18     6-i. 
5 

Prod. 

34     0 

6 

8 

6. 
D.    c. 
4     75 

6 

3 

13. 

6. 
s. 
12 

d. 
11 

7 

s 

i 

£ 

31 

7. 

s.      d. 
16     8J- 
8 

m. 

5 
12 

E, 

2 

8. 
D.    d.  c.  m. 

7     8     9      1 

9. 
£     s.    d, 

4     13     4; 
10 

£ 
8 

10. 

s.       d. 
15     11^ 
11 

11. 
D.     c. 

35     87 

12. 
£      s.       d. 
14      17       8A 

9 

133     19     2-1. 

In  the  last  example,  I  say,  9  times  1  is  9  farthing8=2id.  I  set 
down  \  and  carry  2,  saying,  9  times  8  is  72,  and  2  1  carry  makes 
74  pence, =6s.  2d.  1  set  down  2  in  the  pence  and  carry  6  ;  then, 
9  times  7  (the  unit  figure  in  the  shillings)  is  63,  and  6  I  carry  is  69, 

*  The  produ(Sl  of  a  number,  confifting  of  fcvcral  parts  or  denominations, by 
any  liraple  number  whatever,  will  be  exprcflcd  by  taking  the  proJudt  of  that 
fimple  number,  and  each  part  by  itfclf,  as  fo  many  diftintSt  quefHons :  Thus 
^eSri  15s.  *jd.  multiplied  by  5,  will  be  \k)'o  iFrs.  45d.— (by  taking  the  Oiillings 
from  the  pence,  and  the  pounds  from  the  fliillings,  and  placing  them  in  the 
fhillings  and  pounds  rcfped:ively,)  £168  18s.  Cd.  and  this  will  be  true  when  the 
multiplicand  is  any  compound  number  whatever. 


204  COMPOUND  MULTIPLICATION. 

I  set  down  9  under  the  unit  figure  of  the  shillings,  and  carry  6, 
.saying,  9  times  1  is  9,  and  6  i  carry  is  15,  then  I  halve  16,  which 
is  7  and  1  over,  which  I  set  in  the  ten's  place  in  the  shillings,  and 
that,  with  the  9,  makes  19  shillings ;  then  I  carry  the  7  as  pounds : 
Lastly,  9  times  4  is  36,  and  7  I  carry,  are  43  pounds :  I  set  down 
3  and  carry  4,  saying,  9  times  1  is  9,  and  4  1  carry  makes  13,  which 
I  set  down,  and  the  product  is  X133  19s.  2\d. 

Practical  Questions. 

Kote,  The  facility  of  reckoning  in  the  Federal  Money  compare 
^(\  wilh  pounds,  shillings^  &c.  may  be  seen  from  the  examples  in 
this  and  the  following  cases  ;  where  the  same  questions  are  given 
in  both  the  currencies,  as  near  as  can  be,  avoiding  small  fractions. 

In  the  following  examples  in  this  and  the  succeeding  cases,  the 
price  in  pounds,  or  shillings,  &c.  is  in  the  currency  of  New  Jersey, 
Pennsylvania,  Delaware,  and  Maryland,  where  the  dollar  is  7s.  6d. 
in  the  last  example  in  each  case  ;  in  the  last  example  but  one,  the 
price  is  in  the  currency  of  New  York  and  North  Carolina,  where 
the  dollar  is  8s.  ;  and  in  the  other  examples,  in  the  currency  of 
New  England,  where  the  dollar  is  6s.  Thus  in  the  3d  example, 
the  price,  9s.  lOd.  in  the  currency  of  New  York,  &,c.  is  equal  to 
122c.  9m.  ;  and  in  example  4th,  the  price  13s.  7-^d.  =  181c.  7m. 

1.  What  will  9  yards  of  cloth  at  <  j.^^  *  q  *  ?  per  yard,  come  to  ? 

£0  63.  4d.     price  of  one  yard,     '^^c,  9m. 
Multiplied  by  9        yards.  9 


66c.  8m. 


Ans.  £2     8     0     price  of  9  yds.       §8  -00     1 

^     ^         .      ,S    15s.     4d.     I  I  _  S  ^2     6s. 

2.    3  yards  at  ^  ^^  ^^^^   ^^  ^  per  yard  -  J  ^^ 

..      ^                  59s.      lOd.        }  _  J  £2     19s.      I 

•'*                       \p  22c.  9m.  i  "  ■  "  "   ~  ^  ^7  37c.  4m.  J 

\  13s.     7id.      I  _       ^  J  £6  2s.  7id.    } 

^-     ^  "  "  "       l$\  81c.  7m.  5  'I  pQ  35c.  3m.  S     ' 

CASE  IL 

When  the  multiplier,  that  is,  the  quantity,  is  abose  12  :  You  must 
multiply  by  two  such  numbers,  as,  when  multiplied  together,  will 
produce  the  given  quantity. 

Examples. 

1.  What  will  144  yards  of  cloth  cost  at  \  ^^'^    ^^^^  I  per  yard  ? 

£    s.    d.  c.  m. 

0     3     61  price  of  1  yard.       -5764  Or,  -5764 

Multiply  by  12  144  12 

Produces      2     1     6  price  of  12  yards.    23056  69168 

Multiplied  by         12                                 23066  12' 

6764 - 

Answer  £24  18     0  price  of  144  yards. 83'OOIG 

Ans.  1^83-00  u; 


COIWPOUND  MULTrPLICATION.  105 

Questions.  Answers, 

o      o^         .          C63.     3|d.  }                .       ^  £7   lis.  6d.    } 

2.     24  yards  at  J  ^^  ,  ^^^   ^m.  ^  P"  ^"^^  =  }  $25  24c.  8m.  I 

.^       -,«                   5  9s,      lOd.  /                        \  £13  53    6d.    } 

J  ,?1   22c.  9m.  S    ^  '  '  '   '     }p3  18c.  3m.  S 

.       ..                   \  13s.     7id.  i                        ]  £29   19s..  6(\  I 

^  iJl   81c;7m.  ^    *  -  -  -  -     ^  ^79  94c.  8m.  S 

CASE  III. 

When  (he  quantity  is  such  a  number,  as  that  no  two  nitmbers  in  the 
table  will  prodnce  it  exactly  :  'i'hen  multiply  by  two  such  numbers 
as  come  the  nearest  to  it;  and  for  the  number  wanting,  multiply 
the  given  price  of  one  yard  by  the  said  number  of  yards  wanting-, 
and  add  the  products  together  for  the  answer  ;  but  if  the  product 
of  the  two  numbers  exceed  the  given  quantity,  then  find  the  value 
of  the  overplus,  which  subtract  from  the  last  product,  aud  the  re- 
mainder will  be  the  answer. 


Examples. 
rds  < 

£    s.  d. 


1.  What  will  47  yards  of  cloth,  at    I  173.     9d.        I     ^^  ^,^, 
co*ne  to  ?  I  g2  95c.  8m.  |  ^^^  ^^^^' 


0  17  9  price  of  1  yard. 
Multiplied  by             6 

Produces  4     8  9  price  of  6  yards. 
Multiplied  by             9 

Ans. 

g2-958 

47 

20706 
11832 

Produces  39  18  9  price  of  45  vards. 
Add       1   15  6  price  of  2  yards. 

gl39-02C 

Ans.  £41    14  3  price  of  47  yards. 
Xote.     This  may  be  performed  by  first  finding  the  value  of  48 
yards,  from  which  if  you  subtract  the  price  of  1,  the  remainder 
will  be  the  answer  as  above. 

Questions.  Answers. 

2      75  yards   at  J  ^''  ^^'^'       \  oer  yard  =  \  ^^^   ^'-  ^^'^- 1 
^.     75  yard.,  at  ^  ^3^    ^^^     ^  per  yard  -  ^  ^^^  3^^^  gim.  \ 

.,      ^^,  \  16s.  4d.  I  _  \  £55  2s.  6d.  \ 

^'     ^^^  '  '  '  '    }f,2  4c.  ^  -  '  "  '   -^gi37  81c.  i 

.        Q  ^  \0i.     Od.  i  i  £29   10s.  Od.  I 

"  "  ■  "   <  gl  23cf  S  '   '  "  ■    "^  <  S^S  66c.  6m.  S 

CASE.  IV. 

When  the  quantity  is  any  number  above  the  Mulfiplicoiinn  TabU : 
Multiply  the  price  of  1  yard  by  10,  vviiich  will  produce  ihe  price 
of  10  yards:  This  product,  multiplied  by  10,  wiji  give  the  price  of 
100  yards  ;  then,  you  must  multiply  the  price  of  ^-r.ft  hundred  by 
the  number  of  hundreds  in  your  question  ;  (he  price  often,  by  liie 
number  of  tens  ;  aod  the  price  of  unity,  or  1,  by  the  number  of 

O 


lOe  COMPOUND  MULTIPLICATION. 

units  :  Lastly,  add  these  several  products  together,  and  the  sum 
will  be  the  answer. 

Examples. 

1.  What  will  369  yards  af  cloth,    at  5  4s.   7|d.  }  , 

mount  to  ?  I  lie.  Im.  \  P^^  ^^^^'  *' 

-£  s.  d.  c.  m. 

0  4  7i  price  of  1  yard.  -771 

— 359^ 

10  


■ — 6939 

2  6  3  price  of  fO  y^rds.  3855 

10  2313 

23  2  6  price  100yds.  Ans.  g276-78^ 

% 

69  7  6  price  of  300  yards. 
5  time*  the  price  of  10  yds.  =i  11  11  3  price  of  50  yards. 
S  times  the  price  of  1  yd.=  2  1  71  price  of    9  yards. 

Answer  JGSS  0  4|  price  of 359  yards. 

t  297  yards  at  \  ^^''  ^"^^'       ?  per  yard  =  \  f  ^'^^  ^^''  '^^- 1 
z.  z\ii  yaras  at  ^  ^^  gg^   ^^^   ^  per  yara       ^  ^^^  ^^^    ^^  ^ 

J  9s.    Hid.         >  _      .  .  .^^  £235  Os.Sid.     } 

?  §1  65c  61m.  i  ^  I  ^783  40c.  6im.  \ 

i  5s.  lOd.           >  _^  £149  6s.  8d.      \ 

^-  ^1^  -  -  ■     \  $0  72iic.       5 I  $313  331c.  < 

^    _^               ^  18s.  9d.           >  _  <  £717  3s.  9d.      > 

^'  ^^^  '  "  "     ^  ^2  50c.           I  .  -  -  -  ~  ^  gjgj2  50c.          I 

CASE  V. 

To  find  the  value  of  one  hundred  th eight :  As  112  is  the  gross 
hundred,  so  112  farthings  are  —2s.  4d.  and  112  pence  — 9«.  4d.  ; 
therefore,  if  the  price  be  farthings,  or  not  more  than  3d.  multiply 
2s.  4d.  by  the  farthings  in  the  price  of  1  lb.  or,  if  the  price  he 
pence,  multiply  9s.  4d.  by  the  pence  in  the  price  of  I  lb.  aud  in 
either  case,  the  product  will  be  the  answer. 

Examples. 

1.  What  will  1  Cwt.  of  chalk  come  to  at  J'g^^l'^   >  per  pound? 

112  farthings  =  024  price  of  1  Cwt.  at  }  per  lb.  -021 

J^d.  =  6  farthings  in  the  price.  112 

AiisvTcr  £0  14     0  price  of  1  Cwt.  at  li  per  lb.  42 
....                                     ■'                        21 

21  "^-* 
« 

Ads,  2-352 


COMFOUND  MULTIPLICATION.  1^7 

8.  d. 
2.   1  Cwt.  of  tin  at  SJd.  per  lb.  ?  2  4  price  of  1  Cwt.  at  id.  per  lb. 
•03125  9  Jarthings  in  the  price  of  1  ib. 


112 

625.0 
3125 
3125 


Ans.  ^l   1  0  price  ©f  1  Cwt.  at  2i  per  lb. 
f 


8.    d. 


Ans.  <g3-50000 

3.  1  Cwt.  of  lead  at  ^oJ"!*      ?  Ib.  ?  9  4  pr.of  iCwt.atld.  perlb. 
(tfc.tjm.  )  7  pence  in  the  price  of  lib. 


Ans.  glQ-976. 


pence  in  me  price 
£S  5  4pr.ofl  Cwt.at7d.perlb, 


Questions.  Answers. 

J.'  1  Cwt.  of  copper  at  0|d.  per  lb.  ='£0  7s. 
r       ,  S    3d.    >         |.         ^  £1  8s.   > 

^-     ^ ^  5c.    \ l$o  60c.  I 

CASE  VI. 

To  ^n<i  the  value  of  a  hundred  weight,  when  the  price  of  lib.  is 
any  number  of  pounds  and  shillings;  or  shillings,  pence  and  far- 
things:  Multiply  the  price  of  1  Ib.  by  7,  its  product  by  8,  and  this 
product  by  2  ;  which  last  product  will  be  the  answer  required  : 
for  7X8X2=112. 

Examples. 

C   5s   7-d.  1 
1.  What  will  1  cwt.  of  tobacco  cost  at  Jgo  '  n]^'  j  per  lb. 


£.  8.  d. 

0  5  71  price  of  1  lb. 

7 

D. 

•9375 
112 

1  19  41  price  of  7  Ib. 

8 

18760 
9375 
Q375 

15  15  0  price  of  56  lb.  or  i  cwt. 

2 

gl05-  Ap 

Ans.  .€31   10  0  price  of  1121b.  or  1  cwt. 

Questions.  Answers. 

01    ^  n    ,     »  V   3s.  iOtd.    7         ..         K      £21    14s.      i 
''  1  Cwt.  at  J  g^Jg^     |perlb.  =  J^,2  3.^    2^,  | 

o       ,  C    9s.     6d.    7  C         £63  4s.        7 

■'*     ^       '  '  I  $\   58ir.  I  -     I     gi77  331c.     i 


10^  COMPOUND  MULTIPLICATION. 

4    ICwt.at?    16s.  Hid.   J  ^C£94  19s.  4d.   j 

^     ,                y  IS-*.   4.1.  i               _  C  €74   13s.  4ii  7 

^*   '   ■  *  '     I  s^l   771c.  5  ■               15186  66|c.  5 

P     J                 C  22s.  Gd.  I               _ci:i26  Os.  Oil  7 

i  ;^3  00c.  i  ■  ■  ■  ~  1^^336  00c.  5 

Practical  Questions  in  Weights  and  Measures. 

1.  What  is  the  weight  of  4  hogsheads  of  sugar,  each  weighing 
7cwt.  3qrs.  191b.  '  Ans.  Slcwl.  2qrs.  201b. 

2.  What  is  the  weight  of  6  chests  of  lea,  each  weighing  3cwt. 
2qrs.  91b.  ?  Ans.  2Icwt.  Iqr.  26Ib. 

3.  If  I  am  possessed  of  li  dozen  of  silver  spoons,  each  weigh- 
ing 3oz.  5pwt. — 2  dozen  of  tea  spoons,  each  weighing  15pwt.  14gr. 
— 3  silver  cans,  each  9oz.  7pwt. — 2  silver  tankards,  each  21  oz. 
15pwt.  and  6  silver  porringers,  each  lloz.  ISpvvt.  pray  what  is 
the  weight  of  the  whole  ?  Ans.  181b    4oz.  3pwt. 

4.  In  35  pieces  of  cloth,  each  measuring  27|  yards,  how  many 
yards?  Ans.  9711  yards. 

6.  How  much  brandy  in  9  casks,  each  containing  45gal.  3qts. 
Ipt  ?  Ans.  412gal.  3qts.  Ipt. 

G.  If  I  have  9  fields,  each  of  which  contains  12  acres,  2  roods 
and  25  poles ;  how  many  acres  are  there  in  the  whole  ? 

Ans,  113A.  3r.  25p. 


COMPOUND  DIVISION^ 

TS  the  dividing  of  numbers  of  different  denominations  :  In  doing 
which,  always  begin  at  the  highest,  and  when  you  have  divided 
that,  if  any  thing  remain,  reduce  it  to  the  next  lower  denomina- 
tion, and  so  on,  till  you  have  divided  the  whole,  taking  care  to  set 
down  your  quotient  figures  under  their  respective  denominations. 

Introductory  T-xamples. 
1.  2.  3 

£       s.     d.  D.     d.    cm.  £      s.     d. 

Divide  549     17     9  by  5         3)14     1      9     6  4)731     5     lOi 

Qnot.  £109      19     6| 


*  To  divide  a  number  consisting  of  several  denominations  by  any  simple 
number  wliatcvcr,  is  the  same  as  dividing  all  the  parts  or  memhers  of  which 
tlvit  number  is  composed,  by  the  pame  number.  And  this  will  be  true  when 
any  of  the  parts  are  not  an  exact  multiple  of  the  divisor  ;  for,  by  conceiving 
the  numbrr,  by  whicli  it  exceeds  that  multiple,  to  have  its  proper  value  by  be- 
ing placed  in  the  next  lower  denomination,  the  dividend  will  still  be  divided 
into  parts,  and  the  true  quotient  foui^d  as  before  :  Thus  x  i  i  !7s.  <  d.  divided 
by  t,  will,  be  the  same  as  £06  ll*7s.  lid.  divided  by  6,  which  is  equal  io  £6 
j;  f.  7d.as  bv  the  rule. 


COMPOUND  DIVISION.  109 


4. 
£    s.    d. 

2)97  19   lOi 

£48   19   iq 

5.                            6. 
£   s.    d.                D.    c.  m. 

6)37   11   4f            7)25  49  4 

7. 
£     s.    d. 
8)739  12  li 

8. 
D.    c. 

10)37  60 

9.                            10. 
£   s.    d.               £    s.     d, 

10)79   13  91-         11)58   19   111 

11. 
E.  D.  d.  cm, 

12)3  9    8,  7    5 

In  the  first  example,  having-  divided  the  pounds,  the  4,  which 
remains,  is  4  pounds,  which  are  equal  to  80  shilling.*,  and  17  in  the 
shillings  make  97  ;  1  then  find  5  is  contained  19  times  in  97,  and  2 
over:  I  set  down  19  under  the  shillings,  and  reduce  the  2  shillings, 
which  remain,  into  pence,  and  they  make  24,  and  the  9  pence,  in 
the  question,  added,  make  33  :  Then  how  often  5  in  33  ;  1  find  it 
6  times,  and  3  over :  I  set  down  6  under  the  pence,  and  reduce  the 
3  pence,  which  remain,  to  farthings,  and  they  make  12  ;  then,  how 
often  5  in  12  ;  I  find  it  to  he  twice  :  I  therefore  set  down  -^d.  and 
the  2  which  remains,  is  f  of  a  farthing,  which  I  make  no  account 
of. 

12  13  14  15 

T.  cwt.  qr.  Ife  oz.  dr.  T.  cwt.  qr.  ft  cwt.  qr.  ife  ft  oz.dr. 
3)29    13    2    25    12    13     4)6    11    3    19     5)14    1    12     6)10   13  d 


16.  17.  18.  19. 

ft    oz        ftoz.pwt.gr.       ft  oz.pwt.gr.  ft     oz.   pwt.gr. 

)20    13     8)7    10    15    2     9)56    9    13    19     10)849     11     12    14 


20.  21. 

M.  w.  d.     h.    m.  M.     d.     h.     m.  22. 

6)6     3     5     10     29  7)9     21      12     45         ,8)3s    25°  55'  25'^ 


•     23.  24. 

9)1&°     45'     38"  ]2)18C°     37'     29'' 


25.  Suppose  thai  two  places  lie  east  and  \vest  ol  eacti  other,  and 

it  is  found  by  observation  that  it  is  noon  at  the  form'^r  2  hours,  6' 

30"  sooner  than  at  the  latter  ;  how  many  degrees  are  they  asunder^' 

4')2h.  6'  30"  Keduce  the  hours  and  minutes  to  min- 

,_ .  utes,  then,  minutes  divided  by  4'  give  de- 

31**  37'  30"  Ans.     grees  in  the  quolient.* 

*  Because  360°,  the  wliolp  circumference  of  the  earth,  divided  by  24,  the 
hours  in  a  day,  give  15°  for  one  hour  or  60  minutes :  and  6o-:-i5=:4'  for  oae 

degree. 


Hit  COMPOUND  DIVISION. 

26.  The  longitude  of  Cambridge  is  4h.  44'  17",  and  that  of  Phi- 
ladelphia, 5h.  0*  43" ;  how  many  degrees  difference. 

4°  6'  30"  Ans. 

Practical  Questions. 

CASE  I. 

Having  the  price  of  any  number  of  yards,  pounds^  ^c.  which  i^ 
iwithin  the  pence  table^  or  is  a  composite  number^  to  find  the  price  of 
one  yard^  pound,  ^'C.  use  the  following  rule.  If  the  quantity  do  not 
exceed  12,  proceed  by  setting  down  the  price,  and  dividing  it  by 
the  quantity ;  which  quotient  will  be  the  price  of  one  yard,  re- 
quired ;  but  if  the  quantity  exceed  12,  then  divide  by  such  num- 
bers, as,  when  multiplied  together,  will  produce  the  quantity,  and 
the  last  quotient  will  be  the  value  of  1  yard,  required. 

Examples, 

1.  If  9  yards  of  cloth  cost  £4=  3s.  7id.  what  is  it  per  yard? 

£     s.     d. 

9)4    3     7i 

0    9     31  Ans. 

2.  If  7  ells  cost  £5  17s.  5d.  what  cost  1  ell?     Ans.  16s.  9id. 

3.  If  11  sbeep  cost  £6  5s.  9d.  what  did  each  cost?  Ans.  lis  5id. 

4.  If  12  gallons  of  rum  cost  J£8  Us.  9id.  what  is  it  per  gallon  ? 

Ans.  14s.  3fd. 

5.  If  84  cows  cost  £253  13s.  what  is  the  price  of  each  ? 

Ans.  £3  Os.  4f  d, 

6.  If  132  bushels  of  corn  cost  £20  12s.  6d.  what  is  that  per  bush- 
el? Ans.  OS.  lid. 

7.  If  11  sheep  cost  g25.63c.  what  is  the  price  of  each? 

Ans.  g2.33G. 
S.  If  84  cows  cost  g863.52c.  what  is  the  price  of  one  ? 

Ans.  ^10.28c. 

9.  If  13^  bushels  of  corn  cost  $QQ  what  is  the  price  of  a  bushel  ? 

Ans.  50c. 
•  Note.     If  there  be  a  remainder  after  the  division  by  one  of  the 
parts  of  a  composite  number  before  the  last,  that  remainder  must 
l«€  divided  according  to  the  rule  for  division  of  fractions,  as  in  the 
following  example 

10.  li"  35  yards  cost  £37  lis  what  is  the  price  of  one  yard? 

£  8. 

35=5X7  5)37  ..11 

7)  7  ,  .  10  .  .  2  .  .  Ifqr. 

1  16       Iffqr.  Ads. 

In  dividing  the  farthings  by  7,  there  is  a  remainder  of6|qrs. 
which  is  to  be  divided  by  7  to  obtain  the  whole  answer.  Now  6^=t 
Y,  and  3/-t-7;5?||,  which  must  be  annexed  to  farthing  in  the  last 
?|uotient. 


COMPOUND  DIVISION.  111^^ 

f'l.  If  42  bnshels  of  wheat  cast  g48.57c.  what  costs  one  bushel? 

Ans.  gl.l3y\c. 
12.  What  do  I  pay  a  pound  for  cotton,  when  99Ibs.  cost  £6  2s.  4d. 

Ans.  Is.  2||d. 

CASE  II. 

Having  the  price  of  a  hundred  weighty  to  find  the  price  o/*  1  Ife  : 

Divide  the  given  price  by  8,  that  quotient  by  7,  and  this  qu«' 
tient  by  2,  and  the  last  quotient  will  be  the  price  of  1  jfe  required. 

EX-IMPLES. 

1.  \^  1  cwt.  of  flax  cost  £2  7s.  8d.  what  is  that  per  fe'* 

8)  £2  7s.  8d. 


f)0     5  11 


2)0     0  lOd.Oiqr. 


0     0     5^3_jj,  price  of  1  pound. 

'i.  At.£3  10s.  per  cwt.  what  cost  life  ?  Ans.  1{^. 

3.  At  £6  6s.  per  cwt.  what  cost  life  ?  Ans  Is    Ud. 

4.  At  £42  lis.  8d.  per  cwt.  what  cost  1  3fe?  Ans.  7s.  7id. 
6.  At  ^5.60  per  cwt.  what  cost  1  Ife  ?  Ans.  be. 

6.  At  g3.33c.3m.  per  cwt.  what  cost  1  Ife  ?  Ans.  2c.  9y\«^. 

7.  If  1  cwt.  cost  $156,  what  is  the  price  of  1  ife  ?  Ans.  gl.3i?fc, 

CASE  in. 

Having  the  price  of  several  hundred  TU'eight^  to  find  the  price  per  ife : 
Divide  the  whole  prke  by  the  number  of  hundreds,  which  will  give 
the  price  per  cwt.  and  then  proceed  as  in  the  last  Case. 

Examples. 

t.  If  5^  cwt.  of  sugar  cost  £13  83.  4d.  what  is  that  per  ife? 
§)£13,  8s.  4d. 

S)  2  13    S  price  of  1  cwt. 

7)  0    6    8^d.  price  of  14  Ife  or  |  cwf. 

2)  0    0  ll}d.  price  of  2  fe  or  ^V  cwt. 

0    0    6|  price  of  1  ife. 

2.  If  8  cwt.  cocoa  cost£l5  178.  4d.  what  is  that  per  Ife  ?  Aas,  4fd 

3.  If  31  cwt.  of  sugar  coat  £9  17s.  2d.  what  is  that  per  Ife  ? 

Ans.  6|k1. 

4.  If  Ifcwt.  of  cotton  wool  cost  £8  lOs.  &d.  whatis  that  pe'"r  ife  ^ 

Aa*f.  U. 


112  COMPOUND  DIVISION. 

•5.  If  3  cwt.  of  raisins  cost  g50.52c.  what  cost  1 

J 
3)50.62 


8)16.84 
7)  2.105 
2)     .300f 

.15c.0y\m.  ^ 

6.  If3|  cwt.  cost  gl7.56c.  what  is  one  Ife  ? 

7.  If  llicwt.  cost  g87.33c>  what  cosi  1  Ife  ? 

8.  If  Sj'cwt  cost  g3.64,  what  cost  1  ft?  Ans.  Ic. 
Note.     This  Case  proves  the  6th  in  Compound  MuUipHcation. 

CASE  IV. 

Having  the  price  of  any  number  of  yards,  ^c.  to  find  the  price  of 
1  yard:  Divide  the  price  by  the  quanUt}'^  beginning  at  the  highest 
denomination,  and,  if  any  thing  remain,  reduce  it  into  the  next, 
and  every  inferiour  denomination,  and.  at  each  reduction,  divide  as 
before,  remembering  each  time,  to  add  the  odd  shillings,  pence,  &c. 
if  there  be  any,  and  you  will  have  the  value  of  unity  rt  quired. 

Note.  If  there  be  j,  |  or  |  of  a  yard,  pound,  &c.  multiply  both 
the  price  and  quantity  by  4,  and  then  proceed  as  above  directed  j 
or,  in  federal  money,  work  by  decimals. 

Examples. 

n  If  db^m  of  figs  cost    J  ^^^  11^  l^^l  j  what  are  they  per  fe  ? 

Ife  £'  's.      d. 

(I^uantity  =:  95J-    -  Price  =  16     13     6f 

Mult,  by         4  4 

Produces   382  for  a  divisor.  Product£66     14     3  for  a  dividend. 

382)66     14  3  (0     3     6f  |f|  per  fe. 

20  $    cm. dec,  c  m. 

95i=:95-5)55-69376(-58  2-f  Ans. 

382)1334(3  "      4775 

,  1146  

7843 

188  7640 

12  ■              

2037 

:^82)2259(5  1910 


1910 
349 


,:^82)  1396(3 
1146 

250 


1275 


COMPOUND  DIVISION.  113 

2.  If  147  bushels  of  rye  cost  £47  12s.  6d.  what  is  it  per  bush- 
el ?  Ans.  6s.  5fd. 

3.  If  331  yards  of  baize  cost  £23  13s.  9id.  what  is  it  per  yard  ? 

Ans.  15s,  5id.  y^%. 

4.  If  147lbs.  cost  gl58  76c.  what  is  the  price  of  1  pound  ? 

Ans.  $1   8c. 

5.  Bought  331yds.  of  cloth  for  g85  63c.  2m. ;  what  did  I  pay  a 
yard  ?  Ans.  $2  57c.  5m. 

Note.     This  proves  the  3d  and  4th  Cases  in  Multiplication. 

Practical  Questions  in  Money. 

1.  Divide   <  ^g. .    r^'    ^      >  among  5  men  and  4  womeD,  and 
give  the  men  twice  as  much  as  the  women. 

Men.  Women.  £  ss.  d.  £    s    d. 

5  and  4   Divide  by  14)273  9  4(19  10  8=1  woman's  share. 


ult.  by2 

14 

4  women. 

10  shares. 
Add     4  women's  shares 

14  the  number  of 
shares  in  the 
=Divi8or. 

133       78 

1              1  <^R         .    ._ 

2  8=women's  share. 

£19 

equal     7 

10  8 

2 

—  £39 
14)149(10 

14       — 

—  £195 
9       78 

12       

1  4=1  man's  share. 
5  men. 

6  8= men's  share. 

2  8=women's  share 

—  £273     9  4  Proof. 

14)112(8   

112 
D 
14)911-555(65.111+  =1  woman's  share. 
84  4  women. 


71  260444= women's  share. 

70  

65-1114- 

15  .  2 

14  

130  222-1-  =1  man's  share, 

15                          5 
14  


—  6511114-  =men's  share. 

15  260-4444-  =vvomen's  share. 


14 


911-5554-  Proof 
P 


Hi  REDUCTION  OF  COINS. 

2.  Divide  J£120  17;?.  4d.  among  7  men  and  7  women,  and  give 
the  women  3  times  so  much  as  the  men. 

Aus.  £4  6s   4d.==a  man's  share.      £12  19s. =a  woman's  share. 

3.  Divide  £39   12s.  5d.  among   4  men,   G  women,  and  9  boys : 
Give  each  man  double  to  a  woman,  and  each  woman  double  to  a  boy. 

Ans.    £1    Is.   5d.=a  boy's    share.      £2    2s.    10d.=:a   woman's 
share.      £4  6s.  8d.=a  man's  ditto. 

4.  Divide  5  guineas  among  8  men  : — Give  A  8d.  more  than  F, 
and  B  8d.  more  than  C,  &c.  Ans.  H's  share=153.  2d. 


REDUCTION  OF  COINS. 

RULES  for  reducing  the  Federal  Coin,  and  (he  currencies  of  the 
several  United  States  ;  also  English,  Irish,  Canada,  Nova-Scotia, 
Livres  Tournois  and  Spanish  milled  Dollars,  each  to  the  par  of  all 
the  others.*  *  "^ 

I.  To    reduce   New    Hampshire^  1.  Reduce   £349   19«.   Id.  to 

Massachusetts  t    Rhode     Island^  dollars. 

Connecticut^  and  Virginia  cur-  '9  =^  the  shillings. 

rency  :  '05  =  odd  shillings. 

1.  To  Federal  Money.  '004  =:  qrs.  in  Id. 

Rule. —  Reduce  the  shillings,        

pence  and  farthings,  to  decimals,  '954  =  decimal, 
by   Inspection    (Case   3d,    Dec.  3)349  954         D.     c.  m. 
Frac.)   divide  ^the   whole  by   3,  1166  513=1166  51   3  Ans. 
putting   the   comma   one   figure  2.  Reduce  19s.  Ifd,  to  dollars- 
further  to  the  right  hand  in  the  ^3  19c.  Ans. 
quotient,  than  in  the  pounds  of  3.  Reduce  Is.  to  cents, 
the    dividend,  and   the  quotient  Is.  =  -05  then 
will    be  the   answer  in  dollars,  3)0  5               c.  m. 
cents  and  mills,  0-  166|  =  16  6|. 

•  The  Rules  for  the  reduction  of  money  depends  upon  the  relative  value  of 
the  currency  of  different  States,  &c.  This  value  is  given  in  the  several  rules. 
Thus  4  pounds  of  the  currency  of  New  York  and  North  Carolina,  are  equal 
to  3  pounds  of  New  England  and  Virginia;  4  of  New  England. and  Virginia, 
are  equal  to  3  of  England  ;  4  of  New  England  &c.  are  equal  to  5  of  Pennsyl- 
vania, New  Jersey,  &c.;  and  i  pound  of  New  England,  &c.  is  equal  to  seven 
ninths  of  a  pound  of  South  Carolina  and  Georgia,  and  thus  of  the  others.  The 
rules  are  therefore  obvious.  In  some  cases,  the  process  is  contracted,  and  the 
contraction  is  given  for  the  rule,  because  the  operation  is  simpl.ified.  Thus 
the  first  rule  is  equivalent  to  multiplying  the  pounds  and  .decimals  by  20  and 
dividing  the  product  by  6,  the  shillings  in  a  dollar.  But  as  2^1=;  li  ,  the  sum 
is  multiplied  by  10  in  removing  the  separatrix  one  figure  to  the  right,  before 
the  division  by  3  is  made.  The  relative  value  of  a  i.', depends  on  the  number 
of  shillings  reckoned  to  the  dollar, — the  greater  the  uumbrr  of  shillings  in  a 
dollar,  the  less  the  £,  and  the  reverse. 


REDUCTION  OF  COINS. 


15 


4.  Reduce  Id. 

Ans.  Ic.  3Am. 

5,  Reduce  1  qr. 

Iqr.  =  -OOlOll  and 
3)'0  01  041 

000  347  =  SyW  »«»*^*- 
2.  To   New    York  and   North 
Carolina  currency. 

Rule. — Add  one  third  to  the 
New  Hampshire,  &c.  sum,  and 
the  sum  total  will  be  the  New 
York,  &c.  currency. 

Reduce  £100  NewHampshire, 
&c.  to  New  York,  kc. 
£ 
3)100 
-f  33    6  8 


4^100 

—  25 

£75  Answer. 
6.  To  Irish  Money. 

Rule. Multiply    the    New 

Hampshire,  &c.  sum  by  13,  and 
divide  the  product  by  16. 

Reduce  £100  NewHampshire, 
&c.  to  Irish  Money. 
100 

4x3-l-the  given  sum. 

400 


£133   6  8  Ans. 

3.  To  Pennsylvania,  New  Jer- 
sey, Delaware  and  Maryland  cur- 
rency. 

Rule. — Add  one  fourth  to  the 
New  Hampshire,  &c.  sum. 

Reduce  £100  New  Hampshire, 
^c.  to  Pennsylvania,  £(c. 
4)100 
H-  25 

£125  Ans.. 

4.  To  South  Carolina  and  Geot' 
gia  currency. 

Rule. — -^Multiply  the  New 
Hampshire,  &c.  sum  by  7,  and 
divide  the  product  by  9,  and  the 
quotient  is  the  answer. 

Reduce  £lOONew  Hampshire, 
&c.  to  South  Carolina,  ^c. 
100 

7  ■ 

9)700 
£77  15  6|  Ans. 

5.  To  English  Money. 

Rule. — Deduct  one  4th  from 
the  New  Hampshire,  &c.  sum 

Reduce  £100  New  Hampshire, 
^c.  to  English  Money. 


1200 
-1-100 


16=4X4)1300 
4)325 

£81  5  Ans.. 

7.  To  Canada  and  Nova  Scotia 
currency. 

Rule. Multiply    the    New 

Hampshire,  &lc.  sum  by  6,  and 
divide  the  product  by  6. 

Reduce  £100 New  Hampshire, 
&c.  to  Canada,  &(c. 
100 
5 

6)500 

£83  6  8  Answer. 

8.  To  Uvres  Tournois. 

Note.  12  deniers,  or  pence, 
make  1  sol,  or  shilling,  20  sols, 
or  sous,  1  livre,  or  pound. 

Rule.- — Multiply  the  New 
Hampshire,  &c.  pounds,  by  17i, 
and  the  product  will  be  livres  : 
Or,  multiply  the  sum  in  shillings 
by  7 :  Divide  the  product  by  8, 
and  the  quotient  will  be  Uvres, 
sous,  &€, 


^ 


116 


REDUCTION  OF  COINS. 


Reduce  jGlOONew  Hampshire, 
&c.  to  Livres  J  ournois. 
100  Or,  100 

174-  20 


2000 

7 


8)14000 

Ans.  1760  Viv. 


ld.==lsou.  5ideniers 


Ans.   1750  livres. 
17^  sous. 


17i  livres. 


£l 

II.   To  reduce  Federal  Moneys  to 

JVcra    England    and   Virginia 

currency. 

Rule. — Multiply  the  Federal 
money  by  3,  and  if  it  consist  of 
dollars  only,  Cut  oflf  1  figure,  if 
of  cents  also,  cut  off  3,  and  if 
of  mills,  4  figures  at  the  right 
hand  ;  then  reduce  the  figures 
so  cut  off  to  farthings  each  time 
cutting  off  as  ai  first  and  the  left 
hand  figures  are  pounds,  shil- 
lings, &c.  Or,  reduce  them  by 
inspection. 

1.  Reduce  gl  166  61c.  3m.  to 
New  England  currency, 
gem. 
1166-61  3 
3 


£349  963  9 
20 

s.19'0780 
12 


III.  To  reduce  New  Jersey^  Penn- 
sylvania^ Delaware  and  Mary- 
land currency. 

1.  To  JVew  Hampshire,  Massa- 
chusetts, Rhode  Island,  Connecti- 
cut, and  Virginia  currency. 

Rule.  Deduct  one  fifth  from 
the  New  Jersey,  &c.  sum,  and 
the  remainder  will  be  New- 
Hampshire,  &c.  currency. 

Heduce  £lOONew-Jersey,&c. 
to  New-Hampshire,  &c. 
5)100 
—  20 

£80  Answer. 

2.  To  New  York  and  North  Ca- 
rolina currency. 

Rule.  Add  one  fifteenth  to  the 
New  Jersey,  &c.  sum. 

Reduce  £100  New  Jersey,  &c. 
to  New  York,  &c. 
15=3x6)100 

3)20 

-1-  6  13  4-f  giv.  sum. 

£106  13  4  the  Answer. 

3.  To  South  Carolina  andGeor' 
gia  currency. 

hule.  Multiply  the  New  Jer- 
sey, &c.  sum  by  28,  and  divide 
the  product  by  45,  and  the  quo- 
tient is  South  Carolina  &c. 

Reduce  £100  New  Jersey,  &C: 
to  South  Carolina,  &:c. 
100 


•936  =ld.  nearly. 
Or,  18s.       =  double  of  9. 

Is.       =5inthe2d  place. 
Id.  =3-9or4qrs.that 

45= 

4X7=28 

400 
7 

19s.  Id.  Ans. 
2.  Reduce  45  dollars. 

£13  10s.  Ans. 
C.  Reduce  gl2   7c.  to  N.  E 
money.                £3  12s.  504. 

=5X9)2800 

5)311    2   2| 

£62  4  61  Ad8. 

REDUCTION  OF  COINS. 


11' 


4.  To  English  Money. 

Rule.  Multiply  the  New  Jer- 
sey, 4^0.  by  3,  and  divide  the  pro- 
duct by  5. 

Reduce  £100  New  Jersey,  &c. 
to  English  money. 
100 
3 

6)300 

£60  Answer. 

5.  To  Irish  Money. 

Rule.  Multiply  the  New  Jer- 
sey, &c.  by  13,  and  divide  the 
product  by  20. 

Reduce  £100  New  Jersey,  &c. 
to  Irish. 

100 

4x3-i-lhe  giv.  sum. 


by  10,  and  the  quotient  will  be  li- 
vres,  sous,  &c. 

Reduce  JElOONewJerseyj^'C, 
to  Livres  Tournois, 


lOO 

14 

400 
100 


Or 


ld.=:I?f0US. 

bus. 


»I007    id.=iffoi 

ao>    is"i4fot 

3  i-i=i4Hv. 


aooo 
7 


Ans.  1 400UV.  10)14000 

X400 
8.  To  Spanish  milled  dollars 
Rule.     Multiply  the  New  Jer- 
sey, &c.  pounds  by  2|  and  the 
product  will  be  dollars— Or,  mul- 
tiply them  by  8  :   Divide  the  pro- 
duct by  3,  and  the  quotient  will 
be  dollars. — If  there  be  shillings 
in  the  given  sum,  for  every  7s. 
6d.  add  1  dollar  to  the  quotient. 
Reduce  £100  10s.  New  Jer- 


400 

gey,  &c.  to  dollars. 

3 

100            Or  100 

8                       2 

1200 

+  100 

3)800                   200 

100X|=  66| 

20=4X5)1300 

266|       10s.=     H 

10s.  =  lJ-                 

4)260 

268  as  be- 



Ans.  268  dol.                  [fore. 

£65  Answer. 

6.  To  Canada  and  JVova  Scotia 

IV.  To  reduce  New  York  and  K. 

currency. 

Carolina  currency. 

R«i-le.  Deduct  one   third  from 

1.  To  JVew  Hampshire,  Massa- 

the New  Jersey,  &c. 

chusetts,  Rhode  Island,  Connecti' 

Reduce  £100New  Jersey,  &c. 

cut,  and  Firginia  currency. 

to  Canada,  &c. 

Rule.  Deduct  one  fourth  from 

3)100 

the  New  York,  &c. 

^       -^33     C  8 

Reduce  £100  New  York,  &c. 

to  New  Hampshire,  &;c. 

£66  13  4  Ans. 

4)100 

7.  To  Livres  Tournois. 

—25 

Rule.  Multiply  the  New  Jer- 
sey, &c.  pounds  by  14,  and  the 
product  will  be  Livres  Tournois, 
— or  multiply  the  sum  in  shil- 
lifigs  by  7  ;  divide  the  product 


£75  Answer. 

2.  To  New  Jersey,  Pennsylva- 
nia, Delaware^and  Maryland  cur- 
rency. 


118 


REDUCTION  OF  COINS, 


Rule.  Deduct  one  sixteenth 
from  the  New  York,  &c.  sum. 

Reduce  £100  New  York,  &c. 
to  New  Jersey,  &,c. 

16=4x4)100 

4)25 
—  £6     6 


£93  15  Answer. 

3.  To  South  Carolina  and  €leor- 
gia  currency. 

Rule.  Multiply  the  New  York, 
&c.  sum  by  7,  and  divide  the  pro- 
duct by  12:  The  quotient  is 
South  Carolina,  &c. 

Reduce  £100  New  York,&c. 
to  South  Carolina,  kc, 
100 
7 

12)700 


£58  6  8  Answer. 

4.   To  English  Money. 
Rule.  Multiply  the  New  York, 
kc.  sura  by  9  :     Divide  the  pro- 
duct by  16,  and  the  quotient  is 
English. 

Reduce  £100  New  York,  kc. 
to  English  Money. 
100 
9 


100 

6X6-f  thrice  the 

giv.  sum. 

600 

6 

3600 
4-300=100x3 

64=8X8)3900 

8)487   10 

£60  18  9  Ans. 
6.  To  Canada  and  J^ova  Scotie. 
currency.. 

Rule.  Multiply  the  New  York, 
&c.  sum  by  5,  and  divide  the  pro- 
duct by  8. 

Reduce  £100  New  York,  &c. 
to  Canada,  &,c. 
100 


8)500 

£62  10  Ans. 
7.   7\)  Livres  Tournois. 
Rule.  Multiply  the,NewYork» 
&c.  sum  in  shillings  by  21  :   Di- 
vide the  product  by  32,  and  the 
quotient  will  be  livres,  sous,  &c. 
Reduce  £lOO  New  York,  kc. 
to  Livres  Tournois. 

100         Note. 




20 

ld.=l^%sou. 

16=4X4)900 

ls.=13isou. 



2000 

ll.=13}liv. 

4)225 

21 

£o6  5  Answer. 

2000 

# 

5.   To  Irish  Money. 

4000 

Rule.  Multiply  the  New  York, 

kc  sum  by  39 :   Divide  the  pro- 

32=4X8)42000 

duct  by  64,  and  the  quotient  is  l- 

rish. 

4)5250 

Reduce  £100  New  York,  &.c. 



to  Irish  money. 

Ans,  1312^ 

livres. 

REDUCTION  OF  COINS. 


119 


8.  To  Spanish  milled  Dollars 
Rule.  If  the  New  York  sum 
be  pounds  only,  annex  a  cypher 
to  ihem,  then  divide  by  4,  and 
the  quotient  is  dollars  :  But  if  it 
be  pounds  and  shillings,  annex 
half  the  shillings  to  the  pounds, 
and  divide  as  before,  and  the  quo- 
tient is  dollars. 

Reduce  £100  New  York,  &c. 
to  Dollars. 

4)1000 

250  Doll.  Ans. 
Reduce  £100  8s.  to  Dollars. 

4)1004 

251  Dolls.  Ans. 

V.  To  reduce  South  Carolina  and 
Georgia  currency. 
1.   To  JVesy  Hampshire^  Massa- 
chusetts, Rhode  Island,  Connecticut 
and  Firginia  currency. 

Rule.  Multiply  the  South  Ca- 
rolina, &c.  sum  by  9,  and  divide 
the  product  by  7. 

Reduce  £100  South  Carolina, 
&c.  to  New  Hampsliire,  kc, 
100 
9 


7)900 


1 1   5}  Ans. 


£128 

2.  To  New  Jersey  Pennsylva- 
nia, Delaware  and  Maryland  cur- 
rency. 

Rule.  Multiply  the  South  Car- 
olina,  &.C.  sum  by  45,  and  divide 
the  product  by  28. 

Reduce  £100  South  Carolina 
lie.  to  New  Jersey,  &c. 
100 


9X5=45 


900 
5 


28—4x7)4500 

4)642   17   H 


3.  To  New  YorJz  and  North 
Carolina  currency. 

Rule.  Multiply  the  South  Car- 
olina, &c.  sum  by  12,  and  divide 
the  product  by  7. 

Reduce  £100  South  Corolioa, 
&c.  to  New  York,  &c. 
100 
12 

7)1200 

£171  8  Gf  Ans. 

4.  To  English  Money. 

Rule.  From  the  South  Caroli- 
na, &:c.  sum,  deduct  one  twenty- 
eighth. 

Reduce  £100  South  Carolina, 
&.C.  to  English  Money. 
28=4X7)100 

4)14  5  8f 


3   11 


54  from  100. 


£96     8  6f  Ans. 
5.    To  Irish  Money. 
Rule.  Multiply  the  South  Ca- 
rolina, 4*c.  sum  by  117,  and  di- 
vide the  product  by  112. 

Reduce  £100  South  Carolina, 
4*c.  to  Irish. 

100 


12x94-9  times 

[the  giv.sum. 

1200 
9 


10800 
4-100x9=900 


112= 

=4X4X7)11700 

4)1671 

8 

H 

4)417 

17 

1* 

£160  14  3?  Ans. 


£104     9  3 1  Ans. 
C.   To  Canada  and  Nova  Scotia, 
currency. 


120 


REDUCTION  OF  COINS. 


Rule.  Multiply  the  South  Ca- 
rolina, 4*0.  sum  by  15,  and  divide 
the  product  by  14. 

Reduce  JGIOO  South  Carolina, 
4'C.  to  Canada,  ^c. 
100 


14^=2x7)1500 

2)214  5  84 

£107  2   lOf  Answer. 
7.  To  Livres  Tournois. 
Rule.  Multiply  the  South  Ca- 
rolina, 4'C.  pounds  by  22i,  and 
the  product  will  be  livres. 

Reduce  1001.  South  Carolina, 
^•c.  to  Livres. 

100     Note.  ld.=lf  sous. 
22i  ls.=Ulivre. 


— ""  11. 


:22i  livres. 


200 
200 
50 

Ans.  2250  livres. 
8.  To  Spanish  milled  Dollars, 
Rule.  Multiply  the  South  Car- 
olina, ^c.  pounds  by  30,  and  di- 
vide the  product  by  7,  and  if 
there  be  shillings,  turn  them  into 
dollars,  and  add  them. 

Reduce  £100  South  Carolina, 
t^c.  to  Dollars. 
100 

10x3=30 


7)3000 
Dollars  428^.    Note.  4—83. 


VI,   To  reduce  English  Money, 

1.  To  New  Hampshire,  Massa- 
chusetts, Rhode  Island,  Connecti- 
cut, and  Virginia  currency. 

Rule.  To  the  English  sum, 
add  one  third. 

Reduce  £100  English  to  New 
Hampshire,  &c. 
3)100 
+  33  6  8 


£133  6  8  Ans. 
2.   To  New  Jersey,  Pennsyha^ 
nia,  Delaware  and  Maryland  cur- 
rency 

Rule.  Multiply  the  English 
money  by  5,  and  divide  the  pro- 
duct by  3. 

Reduce  £100  English,  to  New 
Jersey,  4^0. 

100 
5 

3)500 


£166   13  4 
3.    To   New  York  and  North 
Carolina  currency. 

Rule.  Multiply  the  English 
money  by  16,  and  divide  the  pro- 
duct by  9. 

Reduce  £100  English,  to  New 
York,  ^c. 
100 


4X4 


400 

4 

9)1600  # 

£177  15  6|  Answer. 

4.  To  South  Carolina  and  Geor- 
gia currency. 

Rule.  To  the  English  money 
add  one  twenty  seventh. 

Reduce  £100  English.to  South 
Carolina,  4-c. 


REDUCTION  OF  COINS. 


121 


27^3X9)100 


3)11     2  2 


50 
40 


-f  3   14  Of 


£103  14  Of  Ans. 

6.  To  Irish  Money. 
Rule.   To  the  English  sum  adu 
one  twelfth. 

Reduce  JGlOOEnglish  money  to 
Irish  Money. 

12)100 
+     868 


£108  6  8  Ans. 

0.  To  Canada  and  Nova  Scotia 
currency. 

Rule.  To  the  English  sum  add 
one  ninth. 

Reduce  £100  English,  to  Can- 
ada, &c. 

9)100 


4- 


11   2  2| 


£111  2  2|  Answer. 

7.  To  Livres  Tournois. 

Rule.  Multiply  the  English 
pounds  by  23i,  and  the  product 
will  be  livres. 

Reduce  £100  English  to  Li- 
vres Tournois. 

100  Note.  ld.  =  I|f'50us. 
23|  Is  =lJ|ivre. 

£l=23ilivres. 

300 

200 
331 

Liv.  sou. den. 

Ans.  23331  Liv.=2333  6  8. 

8.  To  Federal  Money. 

Rule.  Multiply  the  pounds,  or 
pounds  and  decimals  of  a  pound 
by  40  and  divide  the  product  by 
9,  and  the  quotient  will  be  the 
dolls   or  dollars  and  cents. 

1.  Reduce  £50sterling  to  Fed- 
eral money. 


9)2000 

g222.22|cts.  Ans. 

^.  Reduce  £36  lOs.  9d.  sterl- 

ling  to  dollars  and  cents. 

£36  10s.  9d.=  £36.525,  an 

36.523X40      -,^^„„ 

:l_-==;4l62.33icts.  Ans. 

9  •* 

3.  Reduce  £l  sterling  to  Fed- 
eral money.  Ans.  ^4.44fc. 

4.  Reduce  £1003.5  sterling 
to  Federal  money. 

Ans.  4460 
VII.   To  reduce  Irish  Money. 

1.  To  A''ezio  Hampshire y  Massa- 
chusetts^ Rhode  Island^  Connecticut 
and  Virginia  currency. 

Rule.  Multiply  the  Irish  sum 
bv  16,  and  divide  the  product  by 
lb. 

Reduce  £100  Irish,   to  New 
Hampshire,  &c. 
100 

4X4 

400 
4 

13)1600 

£123  1  Gj\  Ans. 

2.  To  New  Jersey^  Fennsylva' 
nia,  Delaware  and  Maryland  cur- 
rency. 

Kule.  Multiply  the  Irish  sum 
by  20,  and  divide  the  product  by 
13. 

Reduce  £100  Irish    to    New 
Jersey,  &c. 
100 


4x5=20 


400 


13)2000(163  16  11^3  Answer. 


^ 


122 


REDUCTION  Oi'  C0iN:5. 


3.  3o  A'cw  York  and  North 
Carolina  currency, 

Kule.  Multiply  the  Irish  sum 
by  64,  and  divide  the  product  by 
39. 

Reduce  -6100  Irish  to  New 
York,  &c. 

100 


8x8=^64 


800 
8 


s. 


39)6400(164  2/^  Answer. 
4.  To  Souih  Carolina  and  Geor- 
gia currency. 

Rule.  Multiply  the  Irish  sum 
by  1 12,  and  divide  the  product  by 
117. 

Reduce  £100  Irish  to  South 
Carolina,  &;c. 
100 


7X4X4=112 

700 

4 

2800 

4 

_, £      S.      d. 

117)11200(95  14  6-pVt 

Ans. 

5.    To  English  Money. 

Rule.  From  the  Irish 

sum  de- 

duct  one  thirteenth. 

Reduce  £100  Irish  to  English 

money. 

£    s.    d. 

13)100(7   13   10-/3 

100     0     0 

—7   13   10/^ 

92     6     Iji-Ans. 

6.  To  Canada  and  JVova  Scotia 
currency. 

Rule.  To  the  Irish  sum  add 
one  thirty  ninth. 

Reduce  £100  Irish  to  Canada, 
kc.  ■    *. 


£  s.  d. 

39)100(2  11  3^% 

100  . 

+2  11  3-/^ 

102  11  3/^Ans. 
7.  To  Livres  Tournois. 
Rule.  Multiply  the  Irish  sum, 
in  pence,  by  70  ;"divide  that  pro- 
duct  by  39,  and  the  quotient  wilt 
be  sous,  which,  divided  by  20. 
will  be  livres. 

Reduce  £100  Iri&b  to  Livre.« 
Tournois. 

1 00X20X1 2=24000d. 
70 

2-0- 

39)1680000(4307|6 

sou. 

Ans.     Livres.  2153  16^1 


ld.==lfisous. 


ls.=21J7_sous. 


£1=21  liv.   lOifsous. 

VI [ I.  To  reduce  Canada  and  No- 
va Scotia  currency. 

1.  To  JVew  Hampshire,  J^fassa- 
chusetis,  Rhode  Island,  Connecti- 
cut,  and  Virginia  currency. 

Rule.  To  the  Canada,  &c.  sum 
add  one  fifth. 

Reduce  £100  Canada,  &c.  to 
New  Hampshire,  &c. 
5)100 
+  20 

£120  Answer. 

2.  To  Nerv  York  and  North 
Carolina  currency. 

Rule.  Multiply  the"  Canada, 
kc.  sum  by  8,  and  divide  the  pro- 
duct by  5. 

Reduce  £100  Canada,  kc.  to 
New  York,  &c. 
100 
8 

5)800 

£1G0  Answer. 


REDUCTION  OF  COINS. 


3.  To  KexR}  Jersey^  Pennsylva- 
nia^ Delaware jUnd Maryland  cur- 
rency. 

Rule.  TotheCanada,  &c.  sum 
add  one  half. 

Reduce  £100  Canada,  ^c.  to 
New  Jersey',  &c. 
2)100 
4-  50 

^150  Answer. 

4.  To  South  Carolina  and  Geor- 
gia currency. 

Rule.  From  the  Canada,  &c. 
sum  deduct  one  fifteenth. 

Reduce  £100  Canada,  kc.  to 
South  Carolina,  &c. 

15=3X5)100 

3)20 

—  6   13  4 

£93     6  8  Answer. 

,5.   To  English  Money. 
Rule.  From   the  Canada,  &c. 
deduct  one  tenth. 

Reduce  £100  Canada,  kc.  to 
Knglish  money. 
10)100 
—  10 

£90  Answer. 
C.    To  Irish  Money. 
Rule.     From  the  Canada,  &i,c, 
deduct  one  fortieth. 

Reduce  £100  Canada,  kc.  to 
Irish  money. 

40)100 

—  2   10 


£97   10  Answer. 

7.   To  Livress  Tournois. 

Rule.  Multiply  the  Canada,  &c. 
pounds  by  21,  and  the  product 
will  be  livres 

Reduce  £100  Canada,  &c.  to 
!'vres  Tournois'. 


100 

7X3=21 

700  ld.  =  l|sous. 

3  ls.=21sous. 

£l=2Hivres. 

Ans.  2100 

8.  To  Spanish  milled  Dollars. 
Rule.  Reduce  the  Canada,  &c. 
sum  to  shillings  :  Divide  them  by 
5,  and  the  quotient  is  dollars. 
Or,  Multiply  the  pounds  by  4, 
and  the  product  is  dollars  :  And 
if  there  be  shillings  turn  them  in- 
to dollars,  and  add  them  to  the 
product. 

Reduce  £100  Canada,  kc.  to 
dollars. 

100  155   15 

20  4 

620 
4-  3=153. 

Ans. 

$623  Ans. 

IX.   To  reduce  Livres  Tournois. 

1.  To  JVew  Hampshire^  Massa- 
chusetts., Rhode  Island,  Connecti- 
cut, and  Virginia  currency. 

Rule.  Multiply  the  livres  by  2: 
Divide  the  product  by  35,  and  ihe 
quotient  will  be  pounds.  Or, 
Multiply  the  livres  by  8  :  Divide 
the  product  by  7,  and  the  quo- 
tient will  be  shillings. 

Reduce    1760  livres  to  New- 
Hampshire,  &c.  currency. 
1750  Or,   1750 

2  8 

35)3500(100Ans.7)14000 

35  

2|0)200|0 


00 


£100  as  bef. 
2.  ToMrc  York  and  J^orth  Ca- 
rolina currency. 


124 


REDUCTION  OF  COINS. 


Rule.     Multiply  the  livres  by 
32:  Divide  the  product  by  21, 
and  the  quotient  will  be  shillings. 
Reduce   13121  livres  to  New 
York,  &LC.  currency. 
1312-5 
32 


26250 
39375 


•20 


21)42000(200|0 

£100  Answer. 
3.  To  JVew  Jersey^  Pennsylva' 
nia,  Delaware  and  Maryland  cur- 
rency. 

Rule.  Divide  the  livres  by  14, 
and  the  quotient  will  be  pounds. 
Or,  multiply  the  livres  by  10: 
Divide  the  product  by  7,  and  the 
quotient  will  be  shillings. 

Reduce    1400  livres  to   New 
Jersey,  &c.  currency. 
1400 
10 
_- — ,  Or, 
7)14000             14)1400(1001. 
14 


20)200j0 


00 


£100  Ans. 
4.  To  South  Carolina  and  Geor' 
gia  currency. 

Rule.  Multiply  the  livres  by 
2,  divide  the  product  by  45,  and 
the  quotient  will  be  pounds.  Or, 
deduct  one  ninth, and  the  remain- 
der will  be  shillings. 

Reduce  2250  livres  to  South* 
Carolina,  &c.  currency. 
2250  Or, 

2  9)2260 

£  —  250 

45)4600(100  Ans.         

45  2|0)2000lO 


00  JClOOasbef. 

5.   To  English  Money.  * 

Rule.     Multiply  the  livres  by 

G:   Divide  the  product  by  7,  and 

the  quotient  is  shillings  :  Or,  de- 


duct one  seventh  from  the  livres, 
and  the  remainder  will  be  shil- 
lings. 

Reduce  2333ilivres  to  English 
money. 

Or, 
7)2333^ 
—  333i 

2l0)200|0 

£lOOasbef. 


2333-1- 


7;  14000 
2|0)200|0 


Ans.  £100 
6.   To  Irish  Money. 
Rule.     Reduce   the  livres  to 
sous,  then  multiply  them  by  39  : 
divide  this  product  by  70,  and  the 
quotient  will  be  pence. 

Reduce    2153  Hv.  16i|  so.  tc 
Irish  money.  20 

43076|| 
39 


387720 
129228 


710)168000|0 
12)24000 

2j0)200|0 

£100  Answer. 

7.  To  Spanish  milled  Dollars, 
or  to  Federal  Dollars. 

Rule.  Multiply  the  livres  by 
4  :  Divide  the  product  by  21,  and 
the  quotient  will  be  Spanish  or 
Federal  Dollars. 

Reduce  1000  livres  to  dollars. 

1000  Or,  1000 

4  4 

21)4000(190  \  ^p;;;2i)4ooo(i9p  \  j^'j; 


21 

190=19011 
$189 

10 
6 
—  s.  d.  q. 
21)60(2  10  1 


21 

190 
1G9 

10 
10 


d.  c.  m. 

2 


REDUCTION  OF  COINS. 


125 


X.  To  reduce  Spanish  milled  Dol- 
lars. 

1.  To  JVew  Hampshire,  Massa- 
chusetts^ Rhode  Island,  Connecti- 
cut, and  Virginia  currency. 

Rule.  Multiply  the  Dollars  by 
3,  and  double  the  right  hand  fig- 
ure of  the  product,  for  shillings  ; 
the  left  hand  figures  are  poands. 
Reduce  629  dollars  to   New 
Hampshire,  &c. 
529 
3 

£158  14  Answer. 

2.  To  New  York  and  North 
Carolina  currency. 

Rule.  Multiply  the  number  of 
dollars  by  4 :  Double  the  right 
hand  figure  of  the  product  for 
shillings,  and  the  left  hand  fig- 
ures are  pounds. 

Reduce  529  dollars'  to  New- 
York,  &c. 

4 

£211    12  Answer. 

3.  To  New  Jersey  Pennsylva- 
nia, Delaware  and  Maryland  cur- 
rency. 

Rule.  Multiply  the  number  of 
dollars  by  3,  and  divide  by  8. 

Reduce  629  dollars   to  New 
Jersey,  &c. 
529 
3 

£     s.d. 

8)1687(198  76  Answer. 
Or,  8)1587 


Reduce  529  dollars  io  South 
Carolina,  &c. 

529 
7 


£198f  Ans. 

4.  To  South  Carolina  and  Geor- 
gia currency. 

Rule.  Multiply  the  number  of 
dollars  by  7,  and  divide  by  30. 


3|0)370|3 

£l23if  Answer. 
*5.  To  English  Money,  at  4s. 
6d.  per  dollar. 

Rule.  Multiply  the  dollars  by 
9,  and  divide  by  40. 

Reduce  529  dollars  to  English 
money.  529 

9 

4;0)476|1 

£119^V  Answer. 
6.  To  Canada  and  Nova  Scotia 

currency. 

Rule.   Divide  the  dollars  by  4, 
Reduce  529  dollars  to  Canada, 

&c.  4)529     . 

£1321  Answer. 

7.   To  Livres  Tournois. 

Rule.  Multiply  the  dollars  by 
5},  and  the  product  will  be  li- 
vres. Or,  multiply  them  by  2  J  ; 
divide  by  4,  and  the  quotient  wiH 
be  livres. 

Reduce  100  Spanish  dollars  to 
livres.       100 


500 
100X1=25 


Or, 

100 

21 

4)2100 


Ans.  525  livres.    625  as  bef. 

*  Note,  that  in  England  dollars  are 
Bullion,  that  is,  they  are  bought  and* 
sold  by  weight,  and  their  value  varies 
as  other  articles  of  merchandize. 


Note. 


{  1  Cent  =  l^V  Sous.  ) 
M  Dime  =101  Sous.  > 
(1  Dollar  =    51  Livres.  i 


l^fei  DUODECIMALS. 

DUODECIMALS, 

OR  CROSS  MULTIPLICATION, 

IS  a  Rule,  maile  «pe  of  by  workmen  and  artificers  in  casting  up 
the  contents  of  their  works. 

Dimensions  are  generally  taken  in  feet,  inches  and  parts. 

Inches  and  parts  are  sometimes  called  primes,  seconds,  thirds, 
&c.  and  are  marked  thus;  inches  or  primes  ('),  seconds  ('), 
thirds  ('"),  fourths  (""),  &c. 

This  method  of  multiplying  is  not  confined  to  twelves  ;  but  may 
be  greatly  extended  :  For  any  number,  whether  its  inferiour  de- 
nominations decrease  from  the  integer  in  the  same  ratio,  or  not, 
may  be  multiplied  crosswise  •  and,  for  the  better  understanding  of 
it,  the  learner  must  observe,  that  if  he  multiplies  any  denomina- 
tion by  an  integer,  the  value  of  an  unit  in  the  product  will  be  equal 
to  the  value  of  an  unit  in  the  multiplicand  ;  but  if  he  multiplies  by 
any  number  of  an  inferiour  denomination,  the  value  of  an  unit  in 
the  product  will  be  so  much  inferiour  to  the  value  of  an  unit  in  the 
multiplicand  as  an  unit  of  the  multiplier  is  less  than  an  integer. 

Thus,  pounds,  multiplied  by  pounds,  are  pounds  ;  pounds,  multi- 
plied by  shillings,  are  shillings,  &c.  shillings,  multiplied  by  shillings 
are  twentieths  of  a  shilling  ;  shillings,  multiplied  by  pence,  are 
twentieths  of  a  penny  ;  pence,  multiplied  by  pence,  are  240ths  of 
3  penny,  &c. 

Rule.* 
},  Under  the  multiplicand  write  the  corresponding  denomina- 
tiwis  of  the  multiplier. 


*■  The  reason  of  this  rule  is  evident  by  considering^  the  denominations  beluw 
the  integer,  as  fractional  parts  of  the  integer,  and  multiplying  as  in  Vulgar  Frac- 
rions.  Thus  inches  or  primes  aro  12ths  of  a  foot,  seconds  are  12ths  of  an  inch, 
cirl44tlis  of  a  foot,  and  so  on.  Then  feet  multiplied  by  inches  would  give  inchr 
3        6  3        6 


Xj2-^ 


^'\  for  2  feet  X  —=—=6  inches  ;  inches  by  inches  give  seconds^  ^^^Tc) 
18  n  n-{-6  12  6  1        6 

^.o  thirds  ^Ives  fourths  Jor  -^p<  u.^=-^^7^Y:^.f=^j;7^+J^^^ 
r=zl*  and  G"',  and  so  on. 

A  similar  process  will  show  the  correctness  of  the  Rule,  when  the  denomina- 
tions do  not  decrease  uniformly  by  12  or  any  one  number,  as  in  pounds,  shillings 
>.,rA  peace,  where  1  shilling  would  be  _.l.  of  a  pound,  and  1  penny  _J  ,  of  a 
^.ound,  and  so  on. 

Note.     It  is  evident  that  when  the  denominations  decrease  by  any  one  num- 
ber, as  12,  the  denomination  of  the  product  is  the  sum  of  the  denominations  of 
Ihe  factors.     Thus  primes  into  primes  give  seconds,  2  being  the  sum  of  l-f- 1,  the 
■ifraominations  of  the  factors ,  seconds  into  thirds  give  f:i'tl'.s.  2-\-'j^-^i ;  second": 
'jIo  fourths  jive  sixths,  and  io  mi. 


DUODECIMALS.  iST 

2.  Multiply  each  term  in  the  multiplicand,  beginning  at  the  low- 
est, by  the  highest  denomination  in  the  multiplier,  and  write  the 
result  of  each  under  its  respective  term,  observing,  in  duodeci- 
mals, to  carry  an  unit  for  every  12,  from  each  lower  denomination 
to  its  next  superiour,  and  for  other  numbers  accordingly. 

3.  In  the  same  manner  multiply  all  the  multiplicand  by  the 
primes  or  second  denomination  in  the  multiplier,  and  set  the  re- 
sult of  each  term  one  place  removed  to  the  right  hand  of  those  in 
the  multiplicand. 

4.  Do  the  same  with  the  seconds  in  the  multiplier,  setting  the 
result  of  each  term  two  places  to  the  right  hand  of  those  in  the 
multiplicand. 

5.  Proceed  in  like  manner  with  all  the  rest  of  the  denomini^- 
tions,  and  their  sum  will  be  the  answer  required. 

Examples. 


I.  Mnlti 

ply 

H 

feet 

by  2|  feet. 

Or  til 

F.  ' 

2r> 

2     6 

Or  thus. 

2o 

2     6 

21 



21- 

2 

125 

5     0 

60 

1     3 

0 

11 

Ans.  6.25 

Ans.  6     3  — 

Ans.  6i  square  feet  =  6ft.  36in. 

iSo  that  the  3  is  not  3  inches,  hxf. 
36  inches,  or  |  of  a  square  foot, 
2,  Multiply  9f.  8'  6"  by  7f.  9'  3'' 
F.    '    " 
9     8     6 
7     9     3 


67   11     6  ~  Product  by  the  feet  in  the  multiplier. 

7     3     4     6"'  =  ditto  by  the  primes. 

2     6     1     6""   =  ditto  by  the  seconds. 

75     5     3     7     6  Answer. 

3.  How  many  square  feet  in  a  board  17  feet  7  inches  long,  and 
1  foot  5  inches  wide  ?  Ans.  24ft.  10'  11' 

4.  How  many  cubick  feet  in  a  stick  of  timber  12  feet  10  inches 
long,  1  foot  7  inches  ^vide,  and  1  foot  9  inches  thick  ? 

Ans.  35ft.  6'  8''  6"' 

5.  How  many  cubick  feet  of  wood  in  a  load  6  feet  7  inche? 
long,  3  feet  5  inches  high,  and  3  feet  8  inches  wide  ? 

Ans.  82ft.  6'  8"  4" 

6.  There  is  a  house  with  4  tiers  of  windows,  and  4  windows  irr 
a  tier;  the  height  of  the  first  tier  is  6ft.  8';  of  the  second,  5ft.  9' ; 
of  the  third,  4ft.  6';  and  of  the  fourth,  3ft.  10';  and  the  breadth 
of  each  is  3it.  5' ;  how  many  square  feel  do  Ihey  contain  in  th*' 
'^hole  ?  An3,  ^ooft.  7 


J  28  DUODECIMALS. 

The  two  following  questions  are  Sexcesimals. 

7.  If  2  places  diflfer  in  longitude  2**  12' ;  what  is  their  diflerence 
of  lime  ? 

Mult.  £°  12'  00"  00'' 
by  3'  69'  20'"  the  time  in  which  the  sun  passe?  through  1' 

8'  46"  32'"  Answer. 

8.  Two  places  differ  in  longitude  3P  27'  30";  What  is  the  dif 
ference,  in  time,  of  the  sun's  coming  to  the  meridian  of  those  pla- 
ces, the  sun  passing  through  16**  m  an  hour? 

31°  37'  30" 

4'  00"  In  4'  of  a  solar  day,  or  day  of  24  hours,  the  sun  passes  1* 


£«»    6' 30"  00'"  Answer. 

9.  Bought  a  load  of  wood,  which  was  3  feet  wide,  2  feet  8  inch- 
es high,  and  8  feet  long  ;  what  part  of  a  cord  of  wood  did  it  con- 
tain ?  Ans.  Half  a  cord. 

10.  A  load  of  wood  was  4  feet  6  inches  wide,  3  feet  10  inches 
high,  and  7  feet  8  inches  long ;  how  many  feet  more  than  a  cord 
did  it  contain  ?  Ans.  4i  feet. 

11.  A  stick  of  timber  is  1  foot  8  inches  in  depth,  and  2  feet  3 
inches  in  width,  and  42  feet  8  inches  long  ;  how  many  solid  feet 
of  timber  does  it  contain?  Ans.  160. 

12.  Multiply  £3  6  8  by  £2  5  7. 

£     s.    d. 

3     6     8 

2     5     7. 

■ 13.  A,  Band  C  bought  a  drove  of 

i^3X-£2=£6  =600  sheep  in  company  ;  ApaidJ£l46s. 
t>s.XJe2=12s.  =  0  12  0  B,  £13  10s.  and  C, £11  6s.  They 
8d.x£2=16d.  =014  agreed  to  dispose  of  them  at  the 
£3x6s.  =15s.  =  0  15  0  market;  that  each  man  should  take 
l3S.x6s.  =|^s.  =  0  16  18s.  as  pay  for  his  time,  &c.  and  that 
$d.x6s.  =|^d.  =002  the  remainder  should  be  divided  in 
£3x7d.  =21d.  =  0  19  proportion  to  their  several  stocks  : 
6s.X7d.  =f|d.  =  00  2j\  At  the  close  of  the  sale,  they  found 
8d.x7d.  =^Yo<''=  0     0     0^'V  themselves   possessed    of  £46  5s. 

__. what  was  each  man's  gain,  exclu- 

Ans.  7  11      111  sive  of  the  pay  for  his  lime,  &;c. 

£14  6-|-£l3  10-|-£11  6=£39,  and  £46  5— £.39=£7  5,  and 
£7  5~18s.x3=£4  lis.  whole  gain,  and £4  ll"~39=2s.  4d.  gain 
m  the  pound. 

£14     5     0         £13     10  0         £115  0 

X24  X2  4  X24  £s.  d. 


il    13  3 
MUG 


8     6  17  0  12  6       Proof.  .... 

4<>'  46  3  9  (163 


A.£l   13     3        B.£l   11  6        C.£l  6  3  £4  11  0 


SINGLE  RULE  OF  THREE.  12? 


THE  SINGLE  RULE  OF  THREE, 

IS  SO  calleiK  because  three  numbers  are  given  to  find  a  fourth, 
which  shall  have  the  same  ratio  to  one  of  the  given  numbers,  as 
there  is  betwieen  the  other  two.  It  is  usually  distinguished  into 
Direct  and  Inverse.  The  reason  of  this  distinction,  and  the  par- 
ticular rules,  will  be  given  hereafter.  It  will  be  more  easy  how- 
ever, for  the  student  to  proceed  according  to  the  following  General 
Hule  for  stating  and  working  questions  in  the  Rule  of  Three. 

General  Rule* 

1.  Place  that  number,  which  is  of  the  same  name  or  quality  as 
the  answer  sought,  for  the  second  term. 

2.  Consider  whether  the  answer  should  be  greater  or  less  than 
the  second  term.  If  it  must  be  greater,  place  the  greater  of  the 
two  remaining  numbers  in  the  question  on  the  right  for  the  third 

*  This  Rule,  on  account  of  its  j^rent  and  extensive  usefulness,  is  sometimes 
called  the  Golden  Rule  •/  Proportion :  For,  on  a  proper  application  of  it  and 
the  preceding  rules,  the  whole  business  of  Arithnietick,  as  well  as  every  mathe- 
matical enquiry  depends.  The  rule  itself  is  founded  on  this  obvious  principle, 
that  the  magnitude  or  quantity  of  any  effect  varies  constantly  ih  proportion  to 
♦he  varying  part  of  the  cause  :  Thus,  the  quantity  of  goods  bought,  is  in  pro- 
portion to  the  money  laid  out ;  the  space  gone  over  by  an  uniform  motion,  is  iu 
proportion  to  the  time,  &;c.  * 

As  the  idna,  annexed  to  the  term,  proportion^  is  easily  conceived,  the  truth  of 
the  rule,  as  applied  to  ordinary  inquiries,  may  be  made  evident  by  attending  t<"> 
principles,  already  explained. 

It  has  been  shewn,  in  IVlidtiplication  of  Money,  that  the  price  of  one,  multi- 
plied by  the  quantity,  is  the  price  of  the  whole  ;  and  in  Division,  that  the  price 
of  the  whole,  divide<l  by  the  quantity,  is  tlie  price  of  one  :  Now,  in  all  cases  of 
valuing  goods,  Szc.  where  one  is  the  first  term  of  the  proix)rtion,  it  is  plain  that 
the  answer  found  by  this  rule,  will  be  the  same  as  that,  found  by  Multiplication 
of  Money  ;  and,  where  one  is  the  last  ter'm  of  the  proportion,  it  will  be  the  same 
as  that  found  by  Division  of  Money. 

In  like  manner,  if  the  first  terra  be  any  number  whatever,  it  is  plain,  that  the 
product  of  tlie  second  and  third  terms  will  be  greater  than  the  true  answer,  re- 
quired, by  as  much  as  the  price  in  the  second  term  exceeds  tlie  price  of  one,  or 
as  the  first  term  exceeds  a  unit ;  consequently,  this  product,  divided  by  the  first 
lenn,  will  give  the  true  answer  requil-ed. 

Note  1 .  When  it  can  be  done,  multiply  and  divide  as  in  Compound  Multipli- 
cation, and  Compound  Division. 

2.  If  the  first  term,  and  either  the  second  or  third  can  be  divided  by  any 
number  without  a  remainder,  let  them  be  divided  and  the  quotient  used  instead 
of  them. 

The  following  methods  of  operation,  when  they  can  be  used,  perform  the 
work  iu  a  much  shorter  manner  than  tlie  general  rule. 

1.  Divide  the  second  term  by  the  first :  Multiply  the  quotient  into  the  third, 
and  the  product  will  be  the  answer. 

2.  Divide  the  third  term  by  the  first;  multiply  the  quotient  into  the  second, 
and  the  product  will  be  the  annver. 

3.  Divide  the  first  term  by  the  seconil,  and  the  third  by  that  qtiotient,  and  the 
last  quotient  will  be  the  answer. 

4.  Divide  the  first  term  by  the  third,  and  the  second  by  that  qv<otient,  and  the 
laat  quotitat  will  be  the  an^wsr. 

R 


130  SINGLE  RULE  OF  THREE. 

term  ;  but  if  the  answer  must  be  less,  place  the  less  of  the  twc 
numbers  on  the  right  for  the  third  term,  and,  iu  each  case,  place 
the  remaining  number  on  the  left  for  the  first  term. 

3.  Divide  the  product  of  the  second  and  third  terms  by  the  frrst 
term,  and  the  quotient  will  be  the  fourth  term  or  answer  sought. 

Note.  As  all  questions  in  the  Rule  of  Three,  are  readily  solved 
by  this  process,  all  the  statements,  unless  specrally  mentioned,  will 
be  made  according  to  this  rule. 

The  method  of  proof  is  by  invefltng  the  question. 

But,  that  I  may  make  the  method  of  working  this  excellent  Rule 
as  intelligible  as  possible  to  the  learner,  I  shall  divide  it  into  the 
several  cases  following  : 

1.  The  fourth  number  is  always  found  in  the  same  name  iti 
which  the  second  is  given,  or  reduced  to  ;  vvhich,  if  it  be  rrot  the 
highest  denomination  of  its  kind,  reduce  to  the  highest  when  it  can 
be  done. 

2.  When  the  second  number  is  of  divers  dertominalions,  bring  it 
to  the  lowest  mentioned,  and  the  fourth  will  be  found  in  the  same 
name  (o  which  the  second  is  reduced,  whkh  reduce  back  to  the 
highest  possible. 

.3.  If  the  first  and  third  be  of  different  nanffes,  or  one  or  both  of 
divers  denominations,  reduce  them  both  to  the  lowest  denomination 
mentioned  in  either 

4.  When  the  product  of  the  second  and  third  is  divided  by  the 
first ;  if  there  be  a  remainder  after  the  division,  and  the  quotient 
be  not  the  least  denomination  of  its  kind;  then  multiply  the  re- 
mainder by  that  number,  vvhich  one  of  the  same  denomination  with 
the  quotient  contains  of  the  next  less,  and  divide  this  product  again 
by  the  first  number ;  and  thus  proceed  till  the  least  denomination 
be  found,  or  till  nothing  remain. 

5.  If  the  first  number  be  greater  than  the  product  of  the  second 
and  third  ;  then  bring  the  second  to  a  lower  denomination. 

6.  When  any  number  of  barrels,  bales,  or  other  packages  or 
pieces  are  given,  each  containing  an  equal  quantity,  let  the  con- 
tent of  one  be  reduced  to  the  lowest  name,  and  then  multiplie«l  by 
the  given  number  of  packages  or  pieces. 

7.  If  the  given  barrel.-',  bales,  pieces.  &c.  be  of  unequal  con- 
tents, (as  it  most  generally  happens)  put  the  separate  content  of 
each  properly  umler  one  another,  then  add  them  together,  and 
you  will  have  the  whole  quantity. 

ExAMPI.K^. 

I.  If  Glfc  of  sugar  cosi  95.  what  will  SOih  cost  at  the  same  rate  ? 

il'»     s.      Ife 

Here  the  answer  must  be  money,  As  (3  :  9  : :  30  :  the  Answer, 
therefore  9s.  is  the  second  term  ;  as  9 

30115  n.ust  cost  more  than  GJb,  SOlti  •- — 

must  he  placed  on  the  right  of  9s.  for  0)210 

the  thinl  term,  and  CAh  on  the  left  ; 

for  the  fir^t  term.  45s.:=i;^2  5s.An*. 


SINGLE  RULE  OF  THREE.  331 

Again,  By  inverting  the  order  of  the  question,  it  will  be, 
2.  If  9s.  buy  6^  of  sugar,  hojv  much  will  £2  5s.  buy  at  that 
rate?  s.    ife       s. 

As  9  :  6  :  :  45  :  the  Ans. 

^)270(30&  Ang. 

Again,  3.  If  30fe  of  fiugar  be  worth £2  5s.  how  much  may  I  buy 
for  9s.  s.     ife        s. 

A5  45  :  30  :  :  9  :  the  Ans. 

'45)270(6ife  the  Ans. 
270 

Again,  4,  Suppose  £2  5s.  will  buy  30lfe  of  sugar:  What  will  6ife 
of  the  same  sugar  cost?  ife     s.       ife 

As  30  :  45  :  :  6  :  the  Ans. 
.6 

3|0)27|0 

9s.  Ans. 
N.  B.  The  last  Ibree  questions  are  only  the  first  varied^  being 
put   merely  to  show  how  any  question,  in  this  Rule,  may  be  in- 
verted. 
5.  If  5  yds.  pf  cloth  cost  g!0  what  will  2Q  yds.  come  to  ? 
yds.   $       yds. 
As  5  :  10  :  :  20 
10~5==2 

§40  Ans. 
Here  I  divide  the  2d  term  by  the  1st,  and  multiply  the  quotient 
into  the  3d,  for  the  answer. 

yds.  $       yds. 
As  5  :  10  :  ;  20 

4=:20-~5 

g40  Ans. 
Here  I  divide  the  3d  term  by  the  Ui,  and  multiply  the  quotient 
into  the  2d,  for  tlie  answer. 

7.  If  20yds.  cost  ^120,  how  many  yards  may  I  have  for  §30? 
$    yds.      $ 
As  120  :  20  :  :  30 
120—20=^6  quot.  and  30-r-6=5  yards,  Answer. 
Here  I  divide  the  1st  term  by  the  2d,  ahd  then,  the  3d  term  by 
ihe  quotient  for  the  answer.  §     yds.        $ 

Again,  8.  Ai  120  :  20  :  :  30 
120-~30==r4  quof.  an<j   qq^4_-5  yards,  Ans, 


132  SINGLE  RULE  OF  THREE, 

Here  I  divide  the  1st  term  by  the  3d,  and  then,  the  2d  term  hy 
that  quotient  for  the  ansvver. 

9.  If  Icwt.  of  tobacco  cost  £5  12  9^  ;  what  will  8cwt.  ditto  cost? 
cwt.  £  s.  d.     cwt. 
As  1  :  6  12  9i  :  :  8 


Ans.  £45    2  4 
Here  there  is  no  need  of  reducing  the  middle  term,  because  it 
can  be  performed  by  compound  multiplication,  the  first  term  being 
an  unit. 

10.  If  8cwt.  of  tobacco  cost  £45  2  4d  ;  what  is  that  per  cwt.  ? 

£  s.  d. 
8)45  2  4 


Ans.    5  12  9i 


Here  there  is  no  n«ed  of  reducing  the  middle  term,  because  it 
may  be  performed  by  compound  division  only,  the  3d  term  being 
an  unit. 

11.  If  9cwt.  3qr3.  sugar  cost  £27  17s.  6d.  what  will  2cwt.  Iqr. 


1  a  cost? 

£  s.  d. 

2c wt.  Iqr.  life 

9cwt. 

3qrs.             27  17  e 

4 

4 

20 

9 
28 

39 

^57 

28 

12 

— 

_ 

73 

312 

6690 

19 

78 

. 

263 

1092 

!fe          d.           Ife 

As  1092  ;  6690  :  :  263  : 

the 

answer. 

263 

2007 

4014 
1338 

12 

1092)1759470(1611 
1092         - — 

2|0)13|4  3d 

, 

6674        £6   14s. 

3d. 

Answer. 

6552 

1227 

1092 

1350  carried  over. 


SINGLE  RULE  OF  THREE,  133 

Brought  over  1350 
1092 


258 
4 


1092)1 032 (Oqr. 
Note  1.  If  you  look  at  the  staling,  you  will  feee  that  the  first  and 
third  terms  are  of  the  same  kind,  but  of  different  denominations, 
and  therefore  are  reduced  to  the  same  name  or  denomination,  and 
that  the  demand  of  the  question  lies  on  the  3d  term. 

2.  That  the  middle  term,  being  given  in  pounds,  shillings  and 
pence,  is  reduced  to  pence.     Biit, 

3.  If  the  second  term  were  in'federal  money,  it  would  be  suffi- 
cient to  proceed  according  to  decimals.  Thus  :  if  the  price  were 
g92  51c.  7m,  fe     D.  c.  m.     ft 

As  1092  :  92-917  : :  263  :  the  Ans. 
263 


278751 
557502 
185834 


D.c.  m. 

1092)24437-171(22-378-f,  Ant. 
2184 


2597 
2184 


4131 
3276 

8557 
7644 

9131 
8736 

395 
12.  If  57yds.  cost  £69  what  will  9yds.  cost  at  that  rate  ? 
yds.    £    yds. 
As  57  :  69  :  :  9 
9 

67)621(£10  178.  lOd.  2f4qrs.  Ans. 
Here,  all  the  terms  being  whole  numbers,  there  is  do  need  of 
reducing  the  middle  one  until  after  stating. 
The  same  in  Federal  money  would  stand  thus  : 
yds.     D.  c.       yds. 
As  57  :  172-50  :  :  9 

9  D.c.  m. 

57)1552-50(27-23'7 


13 r  SINGLE  KULE  OF  THREE, 

13.  If  my  income  be  109  guineas  per  annum,  I  desire  to  know 
what  I  may  spend  per  day,  so  that  1  may  lay  up  £45  at  the  year'^ 
end?  Ans.£0  5  lOf -J^  per  day. 

Note  1.  You  must  subtract  j£45  from  the  value  of  109  guineas. 
2.    There  being  365  days  in  a  year,  your  question  must  next  be 
stated  thus : 

D.     Quin.      £       D.     s.  d.     qr. 
As  365  :   109—46  ::   J   :  6  10  33^^  the  Ans. 

14.  If  ray  salary  be  <£  13  12s.  6d.  per  annum,  what  does  it  amount 
to  per  week  ? 

Ans.£0  16s.  9^|d. 
The  Stating.  ^     Note.  As  there  are  62  weeks 

W.      .^    -s.  d.     VV.  land  1  day  in  a  year,  you  will  get 

As  52  :  43  12  6  ::   1   :   the  Ans.  Uhe  true  answer  to  the  above 

(  question  by  the  followmg  ratio. 
D.        £    8.  d.      D. 
As  366  :   43   12  5  ::  7  :    16s.  8f|f,K 

15.  Suppose  my  income  to  be  16s.  8||fd.  per  week,  what  is  it 
per  annum  ?  Ans. £43  13s.  7J-  -3^5^. 

Note  1.  You  must  first  reduce  the  middle  term  to  pence. 

2.  You  must  multiply  by  365  (the  denominator  of  the  fraction) 
and  add  to  the  product  the  283  which  remains  ;  and  remember  al- 
ways to  do  so  in  similar  cases. 

3.  You  must  divide  by  7,  the  first  term  and  the  quotient  will  be 
the  answer  in  365ths  of  a  penny,  which  (in  all  similar  cases)  must 
Ibe  first  divided  by  the  denominator,  and  then  brought  into  pounds. 

16.  If  I  am  to  pay  Is.  7d.  per  week  for  pasturing  a  cow  ;  what 
must  I  give  per  week  for  37  cows?  £2   ISs.  7d.  Ans. 

17.  How  many  yards  of  cloth  may  be  bought  for  §195  75c.  of 
which  9-}yds.  cost  gU  2c.?  168yds.  3qrs.  Ans. 

18.  If  I  buy  57  yards  of  cloth  for  49  guineas  ;  what  did  it  cost 
per  ell  EngHsh  ?  £1    10s.    1-^3.  Ans. 

19.  A  merchant,  failing  in  trade,  owes  in  all  £3475,  and  has  in 
money  and  effects  but  £2316  13  4  :  Now,  supposing  his  effects  ar^ 
delivered  up,  pray,  what  will  each  creditor  receive  on  the  pound? 

£  £      s.  d.      £ 

As  3475  :  2316   13  4   ::   1  :£0   13s.  4d.     Ans. 

20.  A  owes  B£3475,  but  B  compounds  with  him  for  135.  4d.  on 
the  pound  ;  pray,  what  must  he  receive  for  his  debt? 

£2316  13s.  4d.  Ans. 

21.  If  the  distance  from  Newburyport  to  York  be  31  miles  ;  I 
demand  how  many  times  a  wheel,  whose  circumference  is  15;}  feet 
will  lurn  round  in  perfoiming  the  journey  ? 

10560  times.  Answer. 

22.  Bought  9  chests  of  tea,  each  weighing  3cwt.  2qrs.  211b.  at 
£4  9s.  per  cvvt.  what  came  they  to  ? 

£147  13s.  8|d.  Ans. 

23.  What  will  37^  gross  of  buttons  come  to  at  13  cents  per 
i'o-tj?  g58  50c.  Aps. 


SINGLE  RUtE  OF  THREE.  135 

^4.  A  farm,  containing  125  A  3r.  27p.  is  rented  at  ^11  50c.  per 
acre  ;  what  is  the  yearly  rent  of  tliat  tarm  ?  'I  ;    ,     < 

5144X  Go.  5fm.  An?. 

25.  If  a  ship  cost  £537  what  are  |  of  her  worth  ? 

Eioh.     £      Eigh.   £>    s.  d. 
As  8  :   537  ::   3  :  201   7  6  Ans. 

26.  If  j\-  of  a  ship  cost  gll63  what  is  the  whole  worth? 

§2058  28c.  5ai.  An5J. 

27.  Bought  a  cask  of  wine  at  7Gc.  5fn.  per  g^allon,  for  §125: 
llow  much  did  it  contain  ?  163o;al.  Iqt.  1  ,\pt.  Ans. 

28.  What  comes  the  insnrance  of  JC537  155.  to,  at£4i  per  cent- 
nm?  £       £         £    s.      £  s.  d. 

As  100  :  4}  ::  537  15  :  24  3  111  ^\  Ans. 

29.  What  come  the  commissions  of  £785  to  at  3^  guineas  per 
cent?  £38  9s.  3^, /-d.  Ans. 

30.  A  merchant  honght  9  pachages  of  clolh,  at  3  gtiineas  for  7 
yards:  each  package  contained  8  parcelts,  each  parcel  12  pieces,  and 
each  piece  20  yards  ;  how  many  dollars  came  the  whole  to,  and 
how  many  per  yard  ? 

Vds.  guin.  pack.  § 
As  7  :  3  ::  9  :  34560  Ans.  for  the  whole  cost. 
Yds.  guin.  yd.    § 
As  7  :  3  ::    1:2  Ans.  per  yard. 

31.  A  merchant  honght  49  tuns  of  wine  for  §910  ;  freight  co«t 
|90  ;  duties  §40 ;  cellar  §31  G7c.  ;  other  charges  §50  and  he 
would  gain  §185  by  the  bargain  ;  what  must  I  give  him  tor  23  tuns  " 

'I'un^-       $       $        $      ^$    ^-      $       $     ^'"n«.     $ 

As  49  :  9 10+90-f  40-4-31  67-f  50-f  185  ::  23  :  613  33c  An^. 

32.  If  §100  gain  §6  in  a  year,  what  will  §475  gain  in  that  time  " 

Ans.  §28  50c. 

33.  The  earth  being  360  degrees  in  circumference,  turns  round 
on  its  axis  in  24  hours  ;  how  far  does  it  turn  in  one  minute,  in  the 
43(1  parallel  of  latitude  ;  the  tlej^ree  of  longitude,  in  this  latitude, 
being  about  51  statute  miles?    II.        D.        M.      M.      M. 

As  24  :  360  X  51    ::    1    :    12^3  Ans. 

34.  Shipt  for  the  West  Indies  225  quintals  offish,  at  15sv  6d.  per 
quintal  ;  37000  feet  of  boards,  at  8i  dolls,  per  1000;  12000  shin- 
gles, at  1  guin.  per  1000  ;  19000  hoops  at  §1^  per  1000,  and  53  half 
joes  ;  and  in  return,  I  have  had  3000  galls,  of  rum,  at  Is.  3d.  per 
gallon  ;  2700  gallons  of  molasses,  at  5^4.  per  gallon  :  ISOOife  oi 
coffee,  at  8^4.  perih;  and  19cwt.  of  su^ar,  at  12s.  34.  percwt.  and 
my  charges  on  the  voyage  were  £37  12s.  pray,  did  I  gain  or  lose, 
and  how  much  by  the  voyage  ?  Ans.  lost £134  9s.  9d. 

35.  If  a  statr,  4  feet  long,  cast  a  shade  (on  level  ground)  7  feet ; 
what  is  the  height  of  ihat  steeple,  whose  shade,  at  the  same  time, 
measures  198  feet  ?^  F.sh.  F.hei.  F.sh.  F.hei. 

As  7:4::    190  :    113}  Ans. 

*  As  the  ra*ys  of  lij;ht  from  the  sun  may  be  considered  parallel,  the  lengths  o: 
the  shadows  must  be  proportioned  to  the  hrigh/t  cf  tho  ob'f>r'<.     H'-ii?^  th"  r.-a 
son  of  the;  statement  of  this  question. 


U^Q 


SINGLE  RULE  OF  THREE. 


*36.  Suppose  a  tax  of  ^755  be  laid  on  a  town,  and  the  intentdry 
of  all  the  estates  in  the  town  amounts  to  ^9345,  what  must  A  pay 
^vhose  estate  is  gl49? 

$  $  $        $?  .  c.  m. 

As  9345  :  755  ::   149  :   12   12  7  Ans, 


*  It  may  not  be  amiss  to  show  the  general  method  of  assessing  town  or  parish 
tiaxes.  Y'irst,  then,  an  inventory  of  the  v^lue  of  all  the  estates,  both  real  and  per- 
sonal, arid  the  number  of  polls  for  which  each  person  is  rateable,  must*  be  taken 
in  separate  columns  :  The  most  concise  way  is  then  to  make  the  total  value  of 
the  inventory  the  first  term,  the  tax  to  be  assessed  the  second,  and  $1  the  third, 
and  the  quotient  will  show  the  value  on  the  dollar  :  2dly,  make  a  table,  by  mul- 
tipliplying  the  value  on  the  dollar  by  1,  %  3,  4,  &;c.^-3dly.  From  the  inventory 
take  the  real  and  personal  estates  of  each  man,  and  find  them  separately  in  the 
tUble,  which  wul  shew  you  each  man's  proportional  share  of  tho  tax  for  real  and 
personal  estates. 

JVote.  If  any  part  of  the  tax  is  averaged  on  the  polls,  or  otherwise,  before  stat- 
ing to  find  the  value  on  the  dollar,  you  must  deduct  the  sum  of  the  average  tax 
from  the  whole  sum  to  be  assessed  :  for  which  average  you  must  have  a  sepa- 
!  ate  column,  as  well  as  for  the  real  and  personal  estates. 

EXAMPLK. 

Suppose  the  General  Court  should  grant  a  tax  of  <f  500000,  of  which  the  town 
of  Nowburyport  is  to  pay  $5312  50c.  and,  of  which  the  polls,  being  15.)0,  are  to 
pay  $1  25c.  each : — The  town's  inventory  amounts  to  ^450000,  what  will  it  be 
<m  the  dollar,  and  what  is  A's  tax.  Whose  estate  (as  by  the  inventory)  is  as  fol- 
I(*wg,  viz.  real  $1376,  personal  $1149,  and  he  has  :^  polls  ? 
Pol.   $   c.        Pol.         $      c. 

Fir?t,  As  1  :  1  25  ::  1550  :  1937  50  the  average  part  of  the  tax  to  be  de- 
ducted from  $5312  50c.  and  there  will  remain  ^"3375. 

Secondly,  As  450000  :  3375  ::  1  :  7im.  on  the  dollar. 


TABLE. 

$    s 

c. 

m. 

$  .    $ 

c. 

m. 

$        $ 

c. 

1  is  0 

0 

n 

20  13  0 

15 

0 

200  is  1 

50 

2  —  0 

1 

5 

30  —  0 

22 

5 

300  —  2 

25 

3-0 

2 

«5H 

40  —  0 

30 

0 

^  400  —  3 

00 

4-0 

3 

0 

60  —  0 

37 

5 

500  —  3 

75 

5-0 

3 

u 

60  —  0 

45 

0 

600  —  4 

50 

6-0 

4 

5 

70-0 

52 

5 

700  —  5 

25 

7-0 

5 

2h 

80-0 

60 

0 

noo  —  6 

00 

8-0 

6 

0 

90  —  0 

f 

5 

900  —  6 

■75 

9-0 

0 

n 

100  —  0 

75 

1000  —  7* 

50 

10  —  0 

7 

5 

IVow,  to  find  what  A's  rate  will  b 
His  real  estate  being  $1376, 1  find  by 
•the  table,  that  $1000  h     -     $7  50c. 
1hat$:300i»       -         -         -     2  25 
that  $70  ii         -         -         -         52  5m 
:indthat$6is   -         -  4  5 

•or  his  real  c^itite  if,lO  32 

In  like  manner  I  find  his  tax 
for  personal  estate  to  be 


Real. 
,1;    c.  m. 

Ptn-!50u.ti. 
.*    c.  m. 

ii  61    71 

Polls. 
?•  c.  m. 

Total. 

;^s    c.  m. 

10  32  0 

3  75 

■iz  60  7/ 

IIJ  < ;?  T>o!: 


,11  25o.  03  ch  fre 


3 


7i 


75 


-f$10  2.2=:r$22  68c.  7im. 

or,  $22  6Cc.  An« 


SINGLE  RULE  OF  THREE.  137 

31.  If  50  gallons  of  water,  in  one  hour,  fall  into  a  cistern,  con- 
taininsf  230  gallons,  and  by  a  pipe  in  the  cistern  35  gallons  run  out 
in  an  hour  ;  in  what  time  will  it  be  filled?  Ans.  lo^h. 

38»  A  butcher  went  with  £41G,  to  buy  cattle  :  Oxen,  at  £22  each, 
cows  atJ£4,  steers  at  £3  lOs.  and  calves  at  £2  10s.  and  of  each  a 
like  number  ;  how  many  of  each  could  he  purchase  with  that  sum  ? 

Ans-  13  each. 

39.  Said  Harry  to  Dick,  my  purse  and  money  are  worth  3|  guin- 
eas, but  the  money  is  worth  eleven  times  as  much  as  the  purse  ; 
pray,  how  much  money  is  there  in  it?  Ans.  £4  3s.  5d. 

40.*  If  I  of  a  yard  cost  |  of  a  j£,  what  will  -pj  of  a  yard  cost  ? 
A^  f  :  i  ::  tV  •  fX^-^f ^fH^  Answer. 

41.  There  is  a  cistern,  having  four  cocks ;  the  first  will  empty 
it  in  ten  minutes  ;  the  second  in  20  minutes  ;  the  third  in  40,  and 
the  fourth  in  80  minutes  j  in  what  time  will  all  four,  running  to- 
gether empty  it  ? 

i}^)     Cist.  xMin.      C^    )      Cist.  Min.  Cist.  Min. 
'^■^    \li)}    :    1    -    CO  :    /"rj^V  As  111  :  60    ;;    i   ;  5}  Ans. 

\^^J                            \    i)  that  is-r-  :  60  ::  1   :  -pr-=5f 
4  4o  * 

11}  Cist. 

42.  A  and  C  depart  from  the  same  place,  and  travel  the  same 
road  ;  but  A  goes  5  days  before  B,  at  the  rate  of  20  miles  per  day  ; 
B  follows  at  the  rate  of  25  miles  per  day  :  In  what  time  and  dis- 
tance will  he  overtake  A? 

M.     M.    I).     M.  D.  D.  D.  M.      D.      M. 

As  25—20  :  1  ::  20x5  :  20.     And,  As  1  :  25  ::  20  :  500 

43.  If  the  earth  revolves  36(>  times  in  305  days,  in  what  timfj 
does  it  perform  one  revolution  ? 

Ans.  23h.  56'  3"  56"'-f  =  1  Sidereal  day.t 

44.  If  the  earth  makes  one  complete  revolution  in  23h.  56'3"-{-, 
in  what  time  does  it  pass  through  one  degree  ? 

Ans.  3'  55"  20'" 

45.  If  the  earth  performs  its  ditirnal  revolution  in  a  solar  day,;}; 
or  24  hours  ;  in  what  time  does  it  move  one  degree  ?       Ans.  4' 

46.  Sold  a  c;Mgo  of  fiax  j^eed  in  Ireland,  for  jC 1 795  10s.  Iri«h 
money  ;  what  does  that  amount  to,  in  Massachusetts  currency,  £81 
6s.  Irish  being  equiil  to  £100  Massachusetts. 

Ans.  £2209   I63.  lid. 

*  If  the  first  term  of  \he  statement  be  a  Vulg-ar  Fraction,  whether  the  other 
terms  are  or  not,  after  the  first  and  third  terms  are  reduced  to  the  same  dc  nomi- 
nation, invert  the  fjrst  term  as  in  division  of  Vulg'ar  Fractions,  and  the  product 
of  the  three  terms  will  of  course  be  t'ne  answer. 

The  student  should  work  the  questions  in  V'ulg'ar,  or  Decimal  Fractions,  ac . 
cording  as  the  rules  for  fractirtus  require. 

t  A  sidereal  day  is  tlic  space  of  time  which  happens  between  the  departure 
of  a  star  from,  and  its  return  to  the  same  meridian  again. 

I  The  solar  day  is  that  space  of  time  wliich  intervenes  between  <ho  '-mVo 
departijn;^  from  any  one  meridiiiu,  luvl  \\r  I'etnrti  t  ■  tl--  •  iTi!"  c  :ui'\. 


i3a  SliNGLE  RULE  OF  THREE. 

47.  My  correspondent  in  Maryland  purchased  a  cargo  of  flou? 
for  me,  for  X437  that  currency  ;  how  much  Massachusetts  money 
must  I  remit  him,  £125  Maryland  being  equal  to £100  Massachu- 
setts, or  5  Mar.~4  Mass.  Ans.  £349   I2s. 

48.  A  bit!  of  pxchangp  was  accepted  at  Newburyport  for  the 
payment  of  £345  10,  for  the  IrUe  vahie  delivered  in  New  York, 
at£i33|  New  York  currency,  for £100  Massachusetts  ditto  ;  how 
much  money  was  paid  in  New  York  ?  Ans.  £460  13s.  4d. 

49.  When  the  exchange  from  Massachusetts  to  Georgia  is  £83^ 
Georgia  per  £100  Massachusetts,  how  much  Massachusetts  money 
must  be  paid  in  Boston  to  balance  £457  Georgia  currency  ? 

Ans.  £548  8s.  Mass. 

50.  A  merchant  delivered  at  Boston £520  Massachusetts  curren- 
cy, to  receive  £400  in  Philadelphia;  what  was  the  Massachusetts 
pound  valued  at  ?  Ans.  £l  53.  Penn. 

51.  If  I  draw  a  bill  of  exchange  for  £537  lOs.  6(\.  Massachusetts, 
to  be  paid  in  Ireland,  at£l23j^ij  Massachusptts,  per £100  Irish,  or 
16  Mass.  for  13  Irish  ;  for  how  much  iri^h  money  must  I  draw  the 
bill?  Ans.  £436  14s.  9|d  Irbh. 

52.  Suppose  a  bill  is  drawn  in  Ireland,  and  payable  in  Boston, 
for £673  12s.  6d.  Irish  ;  how  much  Massachusetts  money  comes  it 
to,  the  exchange  at£81|  Irish,  Der£100  Massachusetts  ? 

Ans.  £829   Is.  Gj%(\.  Mass. 

The  value  of  any  quantrty  of  silver  in  any  of  the  currencies  of 
the  United  States  may  be  found  by  the  following  proportion. 

As  the  number  of  grains,  contained  in£l,  is  to£l  ;  so  are  the 
grains,  in  any  given  quantity,  to  its  value. 

53.  What  is  the  value  of  life  of  silver  in  Massachusetts  currency  ; 
the  pound,  or  20  shillings,  containing  13931  grains?  £s.  d. 

As  1393|  :  1  ::  5760  :  4  2  8. 

54.  If  fydt  cost  ^J  what  will  40iyds.  come"  to  ? 

Ans.  g59  6c.  2^m. 

55.  If  70  bushels  of  corn  cost£l2|,  what  is  it  per  bushel  ? 

Ans.  3s.  7id. 
5G.   If  jV  of  a  ship  cost  £51,  what  are  y\  of  her  worth  ? 

Ans.  £10   18s.  6fd.  f 

57.  At  g3f  per  cwt.  what  will  9|lfe  come  to  ?      Ans.  olc.  3m. — 

58.  A  person  having  4  of  a  vessel,  sells  f  of  his  share  for 
^lOOOJ;   what  is  the  whole  vessel  worth  ?  Ans.^<j;2026  25c. 

59.  A  merchant  sold  5|  yrieces  of  cloth,  each  containing  12|yd5. 
at  12|r;.  per  yard  ;   what  did  the  whole  amount  to  ? 

Ans.  $8  82|c. 

60.  A  buys- of  B  £560J  bank  stock,  at£85!|  per  cent,  what  comes 
it  to?  '   Ans.  £480  7s.  6kl. 

61.  A  merchant  makes  insurance  upon  a  vessel  and  cargo,  valu- 
ed at £3750  IG-j  at  15^  guineas  per  cent,  what  iloes  the  premium 
amount  to?  "  Ans  £013  18s.  5}d. 

62.  A  merchant  in  Holland  draws  a  bl5l  upon  his  correspondent 
ir>  Boston  for  3750  <lucats  at  3-*.  44d.  :  How  much  Massachusettif 
currency  must  he  receive?  Ans,  £1505  12s.  Gd. 


SINGLE  RULE  OF  THREE.  I3y 

63.  A  gentleman  from  Boston  being  in  England,  where  tlie- 
price  of  silver  is  to  that  of  gold,  as  1  to  15^^,  exchanged  158|lfe 
of  silver  for  gold  ;  on  his  return  to  Massachusetts,  where  the  price 
of  silver  is  to  that  of  gold,  as  1  to  16^f,  a  friend,  wanting  his  gold, 
gave  him  the  value  thereof  in  silver  j  what  weight  of  silver  did 
he  gain  by  the  exchange  ? 

IfeS.    G.     ft  S.    ftG.        G.      S.         G.      ft  S. 
As  15yV  :  I  ::  158i  :  lOJ    As  \  :  15:^f-  ::  lOi  :  162|f.  Ans.  4/2^^^- 

64.  A  merchant  bought  a  number  of  bales  of  velvet,  each  con- 
taining 1291^  yards,  at  the  rate  of  g7  for  5  yards,  and  sold  them 
out  at  the  rate  of  §11  for  7  yards  ;  and  gained  §200  by  the  bar- 
gain 5  how  many  bales  were  there  ? 

Yds.    $     Yds.     $  Sold  5  yards  for  7f  Dollars. 

As  7  :   11   ::   5  :  7f  Bought  5yd3.  for  7   Dollars. 

In  5  yards  gained  ^   Dollar. 

$  Yds.       $         Yds.  Yds.      B.       Yds.        B. 

As  ^  :  6  ::  200  :   1166|,  and,    As  12911  :  {  ::   1166|  :  9  Ans. 

Although  the  method  before  laid  down  be  universally  applicable, 

yet  there  are  other  methods  more  ready  and  expeditious  in  some 

particular  cases. 

Rule  L 
If  the  first  and  third  terms  be  fractions,  and  the  second  a  whole 
number,  reduce  the  first  and  third  to  one  common  denominator, 
then,  rejecting  the  denominators,  make  the  numerator  of  the  first, 
the  first  term,  and  the  numerator  of  the  third,  the  third  term,  and 
work  as  in  whole  numbers. 

If  I  of  a  yard  cost  9s.  what  co§t  y^^  Y^rd  at  that  rate  ? 

f=if  and  T^^il-     Now,  as  13  :  9s.  ::   14  :  83.  4Jd.  Ans. 

Rule  U. 

If  of  the  first  and  third  terms,  one  be  1,  and  the  other  a  fraction  : 

})ut  the  denominator  of  the  fraction  instead  of  1,  and  the  iiumera- 

tor  in  the  place  of  the  fraction,  and  work  as  in  whole  numbers,  as 

before. 

if  1  acre  of  lan4  cost  £12,  what  will  f  of  an  acre  cost  at  that  rate  ? 
Den.         £>       Num.        £  s. 
As     8     :      12     ::     5     :     7   10  Ans. 
If  the  question  were  wrought  with  the  fractions,  it  is  evident  that 
the  denominator  would  belong  both  to  the   dividend  and  divisor, 
and  thus  destroy  each  other.     Then  in  the  example  under  Rule  1. 
the  statement  would  be, 

9xH 
As  Jf  :  9  ::  -'*-  :  the   answer=f|x9xiJ-=— -. 

i  o 

And  under  Rule  H.  the  statement  would  be, 

8x12x5     12x5 


12  ::  #  :  answer=- 


8  8         8 

Whence  the  reason  of  the  rules. 

()5.  If  '625  of  a  yard  cost  £  25,  what  will  4'75yds.  come  to  ? 
Yds.      £      Yds.     4-75X-25    ^       -^    ^^ 
'^"5  :  -SQ  ::  ^75  :   >   ,^^^  -c^l-9=:l    18  Ans. 


140  SINGLE  RULE  OF  THREE. 

66.  If  9-75y(ls.  cost  ^11  25c.  what  will  -Syds.  cost? 

Ans.  57c.  eifm. 

67.  There  is  a  cistern,  which  has  3  cocks  ;  the  lirst  will  empty 
it  in  -25  hour,  the  second  in  -75  of  an  hour,  and  the  third  in  1*5 
hour  :  in  what  time  will  it  be  emptied  if  all  three  run  together? 

H.  Cist.    H.     Cist. 

^  (-25  :    1   ::    1   :  4 

As  ?  -75  :   1   ::    1   :   1-3334- 

(  1-5  :    1   ::    1    :   0-6G7-- 

6  Cist. 
N  Cist.  il.  Cist.  M.  H.  m. 

As  G   :    1   ::    1    :  |=:0  1G7—  =  10  Ans. 
G8.  A  conduit  has   a  cock,  which  will  fill  a  cistern  in   '2  of  an 
hour  :  this  cistern  has  3  cocks  ;  the  first  will  empty  it  in  1-25  hour, 
the  second  in  '025  of  an  hour,  and  the  third  in  '5  hour.     In  what 
time  will  the  cistern  be  filled,  if  all  four  run  together? 

Ans.  Ih.  40m. 
GO.   If  19yds.  cost  $25  75c.  what  will  435-5yds.  come  to. 

Ans.  ;^590  21c.  7^,-m. 

70.  If  345yd3.  of  tape  cost  g5  17c.  5m.  what  will  one  yard  cost? 

Ans.  Oc.  Im.  'd. 

71.  If  I  give  gl2  82c.  5m.  for  675  tops,  how  many  taps  will  19 
mills  buy  ?  Ans.  1  top. 

72.  If  when  wheat  is  <^1  per  bushel,  the  two  penny  loaf 
weigh  9Goz.  what  ought  it  to  weigh  when  wheat  is  gl  25c.  per 
bushel?  Ans.  7oz.  ISpwt.  14grs. 

73.  How  much  in  length,  that  is  8^  inches  broad,  will  make  a 
foot  square  ?  Ans.  l6p^  inches. 

74.  What  number  of  men  must  he  employed  to  finish  in  9  days» 
what  15  men  would  perform  in  30  days?  Ans.  60  men. 

75.  If  9  men  can  btiild  a  wall  in  5  months  by  working  14  hours 
a  day,  in  what  time  will  the  same  men  do  it,  when  they  work  only 
10  hours  a  day  ?  Ans.  7  months. 

76.  How  many  yards  of  carpet,  2|  feet  wide,  will  cover  a  floor, 
which  is  18  feet  long  and  IG  feet  wide  ?  Ans.  34ifyds. 

77.  If  745  soldiers  are  to  be  clothed,  and  each  suit  is  to  contain 
31yds.  of  cloth  If  yd.  wide,  and  to  be  lined  with  shalloon  |yd.  wide  : 
houMuany  yards  of  shalloon  will  be  necessary  ?  Ans.  4097|yds, 

78.  If  a  man  count  100  cents  in  a  minute  for  10  hours  in  a  day  : 
in  how  many  days  will  hp  count  a  million  of  cents? 

Ans.  1G|  days. 

79.  Proceeding  to  coimt  at  the  same  rate  as  in  the  last  question  ; 
how  many  men  must  be  employed  for  100  years  of  3G5  days  each, 
to  count  niie  trillion  ?.  "  Ans.  45GG21 004^^1  m^n, 

80  The  number  of  inhabitants  on  the  earth  is  computed  to  be 
750000000  ;  suppose  they  had  each  counted  one  lor  every  second 
fiom  tlie  creation  to  this  time  or  GOOO  years  of  305  days  each  : 
how  many  would  they  have  counted  ? 

Ans.  14I9J12000  billions. 


SINGLE  RULE  OF  THREE.  141 

SI.  In  a  certain  school,  J^th  of  the  pupils  study  Greek,  ^J^  study 
Latin,  f  study  Arilhmelick,  \  read  and  write,  and  20  attend  to  oth- 
er things  ;  what  is  the  number  of  pupils  ? 

oV+rV+l+i^A,  then  20=/^  and  j\  :  20  ::  if  :  100.  Ans. 

To  find  the  value  of  Gold  in  Massachusetts  currency- 
Prob.  1.  Given  the  weight  of  any  quantity  of  gold,  to  find  its 


value. 

Oz.   £ 

Oz.    je    pwt.  s.    gr.    d.                     2| 

Theorem  1 . 

As  1  :  5i  : 

:  12  :  64  ::  1  :  6-»  ::  1  ::  2|(Case  1.)=— 

(Case  2.)=^  (Case  3.)=f,  Therefore, 

Rule  1. — If  the  given  quantity  be  in  grains  ;  say,  As  the  denom- 
inator is  to  the  number  of  grains  ;  so  is  the  numerator  to  their  va- 
lue in  pence. 

1.  What  is  the  value  of  18  grains  of  gold? 

By  Case  1.  By  Case  ?.  By  Case  3. 

Gr.  Gr.  Gr. 

As  1 


18  :: 

^ 

As  2  ; 

:   18   :: 

51 

As  3  :   18  ::  8 

02 

^ 

8 

36 

90 

3)144 

12 

6 



--_ 

— 

48d.=4 

12)48(4s.  Ana.  2)96(48d.=4s. 

Rule  2. — If  the  given  quantity  consist  of  ounces,  pennyweights, 
and  grains,  halve  the  grains,  and  then  proceed  as  in  multiplication 
of  pounds,  shillings  and  pence,  making  the  numerator  in  Case  2d, 
the  multiplier. 

1.  What  is  the  value  of  7oz.  8pwt.  16gr.  of  gold? 
Gr.  gr.         oz.  pwt.  gr. 

16-r-2  =  8,  then,  7     8     8 


^i 

37 
2 

3 

9 

4 

^ 

£39   12  lOf  Ans. 
Rule  3. — If  the  given  quantity  consist  of  poumls  only,  multiply  by 
64,  and  the  product  will  be  the  answer  ;  but,  if  it  consist  of  pound?, 
ounces,  &-c.  it  will  be  most  convenient  to  reduce  the   pounds  to 
ounces,  and  proceed  by  Rule  2. 

1.  What  is  the  value  of  36!b.  of  gold,  at  £64  per  lb.  ? 
64 

144 
216 

£2304  Ans. 


4;^  SINGLE  RULE  OF  THREE. 

2.  What  is  the  value  of  16lb.  903.  12pvvt.  18gr.  of  gold  ? 
12 
—-  pwt.  gr.     gr. 
oz.  189     12    9  ==  18-~e 


948     3    9 
63    4    3 


£1011  ,  8    0  Aus. 
pROB.  2.  To  ascertain  the  value  of  any  given  quantity  of  gold  ic 
Spanish  milled  dollars,  or  federal  money. 

Theorem  2.  Ipwt.  of  gold  —  5Js.  1  dollar  «=  6s.  And, 

"^=f^=|.     Therefore, 

Rule.  Reduce  the  given  quantity  of  gold  to  pennyweights; 
then,  as  the  denominator  is  to  the  given  quantity  ;  so  is  the  nume- 
rator to  the  answer  in  dollars.     Or, 

Divide  by  the  denominator,  and  multiply  the  quotient  by  the  nu- 
merator.     Or, 

Divide  by  the  denominator  and  subtract  the  quotient  from  the 
dividend.     In  either  case,  you  will  have  the  answer. 

1.  What  is  the  value  of  6oz.  6pwt.  of  gold,  in  Spanish  dollars  ? 
20 

pwt.  

As  9  :   126  ::  8  |26  pwt.  Or, 

8  ^^)]26 

Or,  —14 


9)1008  9)126 


112  Ans. 


Ans.      112  Dolls.  14x8  =  112Ans. 

2.  In  7oz.  13pwt.  17gr.  how  many  dollars  ? 
oz,  pwt.  gr. 


7  13 

20 

24 

619 

307 

3689 

^16)29512 

216 

791 

648 

3432 

1296 

SINGLE  RULE  O^  THREE.                     143 

To  find  the  value  of  this  remainder. 

1.  In  shillings,  &c.  2.  In  Federal  Money. 

136  Annex  cyphers,  as  in  division 

6  of  decimals  ;    the    two  quotient, 

places   next   to  dollars,  will  be 

216)816(3s.  cents;  the  third,  mills  ;  the  oth- 

648  ers,  decimals  of  a  mill ;  or  the 

. remainder  with  the  divisor,  will 

168  form  a  fraction  of  a  mill. 
12  216)1360(62c.  9Um 
1296 


216)20]6(9d, 


944  640 

432 


72 


4  2080 

1944 


216)288(liqr. 
216 


113. — 1  " 
2  10         2'i' 


2  16         ¥ 

Prob.  3.    To   ascertain  the  weight  of  gold  equivalent  to  any 
given  sum,  currency. 

Rule  1.  If  the  given  sum  be  in  pence,  reverse  Rule  1.  Theo- 
rem 1.  that  is  ;  As  the  numerator  8  is  to  the  given  sum  in  pence  -. 
90  is  the  denominators  to  the  weight  required,  in  grain«. 
What  weight  of  gold  is  equal  to  4s.  ? 
d.  12 

As  8  :  48  ::  r>  — 

3  4r. 

8)144 

Ans.    18  grains. 
Rule  2.  If  the  given  sum  be  in  pounds,  shillings  and  pence. 

As  1  is  equal  to  y  ;  therefore,  divide  the  given  sum  by  8,  and 
that  quotient  by  2 ;  add  the  two  quotients  together,  double  the  last 
denomination,  and  you  will  have  the  answer. 

What  quantity  of  gold  is  equivalent  to  £45   13s.  4d. 

oz.  pwt.  gr. 

Mark  the  pounds,  shillings  and  ^      8)45     13     4 
pence,  as  oz.  pwt.  and  gr.  ^ 


2)5     14     2>  .  ,, 
2     17     1  ^  ^^'^'^' 


8     11     3+: 


Oz.  8     11     6  Ans. 
Pros.  4.  To  find  the  value  i>f  gold  equivalent  to  ar^y^given  sujr 
in  Federal  money, 


144  RULE  OF  THREE  DIRECT. 

Rule.  As  the  numerator  8  is  to  the  number  of  dollars  ;  so  is  the 
denominator  9  to  the  answer  in  pennyweights  :  Or,  divide  the  dol- 
lars by  the  numerator  8,  and  add  the  quotient  to  the  dividend. 

Or,  divide  as  before,  and  multiply  the  quotient  by  the  denomina- 
tor 9.     In  either  case  you  will  have  the  answer. 

1.  Required  the  weight  of  gold  equal  to  16  dollars. 

As  8  :  76  ::  9  Or  thus,  8)76  Or,  9-ix9=85Apwt. 

9  9i 

8)684  Ans.  85ipwt. 

oz.  pwt.  gr.  

Ans.    85ipwt.=4      5      12 

2.  Reqiiired  the  weight  of  gold  equal  gl59  7uc. 

As  8  :  159  75  ::  9  :  179pwt.  17igr.  Ans. 
9 


8)1437-75 

179-71875 
24 


287500 
143750 

17-25  grains. 


Or,  159-75>^8  -f-  159-75  =  179pwt.  17]gr.  An?. 
Or,  159-75-r-8  X  9  =  179pwt.  17igr.  as  before. 

RULE  OF  THREE  DIRECT  AND  INVERSE. 

Though  Direct  and  Inverse  Proporlion,  are  properly  only  parts 
of  the  same  rule,  yet  for  the  use  of  those  who  may  desire  i(,  the 
common  distinctions  will  be  made  and  the  common  rules  given. 

The  Rule  of  Three  Direct  teaches,  by  having  three  numbers 
given,  to  find  2i  fourth,  which  shall  have  the  same  ratio  to  the  se- 
cond, as  the  third  has  to  the  Jirst. 

The  Rule  of  Three  Inverse  teaches,  by  having  three  numbers 
given,  to  find  a  fourth,  which  shall  have  I  he  same  ratio  to  the  se- 
cond, as  the  first  has  to  the  third.  It  is  also  called  reciprocal  or 
indirect  proportion. 

If  more  require  more,  or  less  require  less,  the  question  belongs 
to  the  Rule  of  Three  Direct.  But  if  more  require  hss,  or  less 
require  more,  the  question  belongs  to  the  Rule  of   riupe  Inverse. 

The  principal  ditficully,  which  will  embarrass  the  learner,  will 
be  to  distinguish  when  the  proportion  is  direct,  and  when  inverse. 
This  must  be  done  by  an  attentive  considf  ration  of  the  qut.'tion 
proposed.  For  more  requires  more,  wlien  the  third  term  is  greater 
than  the  tirst,  and  the  question  requires  the  fourth  term  to  be  greater 
than  the  second  ;  and  less  requires  less,  when  the  third  term  i--  less 
ihau  the  tirst,  and  the  fourth  is  required  to  be  lets  than  the  second. 


RULE  OF  THREE  INVERSE.  145 

More  13  said  to  require  less^  when  the  third  term  is  greater  than 
the  first,  and  the  question  requires  the  fourth  to  be  less  than  the 
second  ;  and  less  requires  mure,  when  the  third  term  is  less  than 
the  first,  and  the  fourth  is  required  to  be  greater  than  the  second. 

RULE  OF  THREE  DIRECT. 
Rule. 

1.  State  the  question  by  making  that  number,  which  asks*  the 
question,  the  third  term  ;  that  which  is  of  the  same  name  or  quality 
as  the  demand,  the  first  term  ;  and  that,  which  is  of  the  same  name 
or  quality  with  the  answer  required,  the  second  terra. 

2.  Divide  the  product  of  the  second  and  third  terms  by  the  first 
term  and  the  quotient  will  be  the  answer. 

JVote.  The  directions  under  the  General  Rule,  as  well  as  the 
demonstration,  apply  to  this  rule. 

Examples. 
X.  If  61bs.  of  sugar  cost  lOs.  what  will  331bs.  cost  at  the  same 
rate  ? 

lbs.     s.       lbs. 
As  6  :   10  ::  33  :  the  answer. 
10 

6)330 

65s. =  £2  15s.  Ans. 
In  this  example  331bs.  asks  the  question,  and  is  made  the  third 
term  ;  6lbs.  being  of  the  same  quality,  is  made  the  first  term  ;  and 
10s.  being  of  the  quality  of  the  answer  required,  is  placed  for  the 
second  term. 

To  invert  the  question,  say, 
s.     lbs.      s.      lbs. 
As   10  :  6  ::  55  :  33  the  Ans. 

2.  If  100yds.  of  cloth  cost  $66  what  will  1  yard  cost? 

Ans.  66c. 

3.  If  my  income  be  Jl75t)  a  year,  and  I  spend  19s.  7d.  a  day, 
how  much  shall  I  have  saved  at  the  end  of  the  year? 

Ans.  £167  12s.  Id. 

RULE  OF  THREE  IJVFERSE,  OR  RECIPROCAL  PROPORTIOA', 

RuLE.t 

State  and  reduce  the  terms  as  in  the  Rule  of  Three  Direct ;  then, 
multiply  the  first  and  second  terms  together,  and  divide  the  product 

*  The  term  which  asks  or  moves  the  question,  has  generally  some  -words  like 
these  before  it,  viz.  What  will?  what  cost?  How  many?  how  long?  how 
much  ?  &;c. 

t  The  reason  of  this  rule  may  be  explained  from  the  principles  of  Compound 
Maltiplicatioa  and  C^mpouiitl  Dinaion,  in  th*  same  manner  as  the  direct  rule.— 


146  RULE  OF  THREE  INVERSE. 

by  the  thinl ;  the  quotient  will  be  the  answer  in  the  same  denonvina 
tion  as  the  middle  term  was  reduced  into. 

If  there  be  fractions  in  your  question,  they  must  be  stated  as  be- 
fore directed,  and  if  they  be  vulgar,  invert  the  third  term  :  Then 
multiply  the  three  terms  continually  together,  and  th6  product  will 
be  the  answer. 

Examples. 

1.  How  mucfi  shalloon,  that  is  f  yard  wide,  will  line  6|  yards  ef 
cloth  which  is  1}  yard  wide  ? 

vd.    yds.  qrs.  qrs.  qrs.    qrs. 

As  *|i  :  6f  ;:  3  As  5  :  27  ::  3 

4  4  6 

5  27  3)136 

4)45 

Hi  yards,  Answi 
The  same  by  Vulgar  Fractions. 
First.   li~f,  6i=\\  and  3qrs.=f.     Then, 

5x27x4 
As  I  :  V  ::  I     And  |xVX|=-^f^^5^-=W=V  =  IUj'^s-  Ans. 

The  same  by  Decimal  Fractions. 
1^=:1.25,  6f==6-75  and  3qrs.  =  '75.     Then, 
As  1-25  :  6-75  ::  '75 
1-25 


3375 
1350 
675 

■'  2.  What   length    of  board    7 

^75)8-4375(1  l-26yds.  Ans.       inches  wide,  will  make  a  square 
7  5  foot? 

In.br.  in.len.  in.br.  in.len. 

As  12  :   12  ::  7|  :   J9i  Ans. 


For  example^  if  4  men  can  do  a  piece  of  work  in  12  days,  in  >vhat  time  will  8  ineu 

"**  " '  As  4  men  :  12  days  ::  8  men  : ^=6  days,  the  Answer. 

And  here  the  product  of  the  first  and  second  terms,  that  is,  4  times  12,  or  48,  is 
evidently  the  time  in  which  one  man  would  perform  the  work.  Therefore,  ii 
men  will  do  it  in  one  eighth  part  of  the  time,  or  6  days. 


RULE  OF  THREE  INVERSE.  U7 

3.  Suppose  I  lend  a  friend  £350  for  6  months,  he  promising  the 
like  kindness  ;  but,  when  requested,  can  spare  but  £125,  how  long- 
may  I  keep  it  to  balance  the  favour  ?  £     Mo.     £     Mo. 

As  350  :  5  ::  125  :  14  Ans. 

4.  Suppose  450  men  afe  in  a  garrison,  and  their  provisions  are 
calculated  to  last  but  5  months  ;  how  many  must  leave  the  garrison, 
that  the  same  provisions  may  be  sufficient  for  those  who  remain  9 
months  ? 

Mo.    M.     Mq.    M.  M. 

As  5  '450  ::  9  :  250,  and  450—250=200  men,  Ans. 

5.  If  a  nian  perform  a  journey  in  15  days,  when  the  day  is  12 
hours  long,  in  hovy  many  days  will  he  do  it,  when  the  day  is  but  10 
hours  ?  Ans.  18  days. 

6.  If  a  piece  of  land,  40  rods  in  length,  and  4  in  breadth,  make 
an  acre,  how  wide  must  it  be,  when  it  is  but  19  rods  long  to  make 
an  acre  ?  Ans.  Crods  6ft.  HyV"- 

7.  If  a  piece  of  board  be  30  inches  in  length,  what  breadth  will 
make  1|  square  foot?  Ans.  7-2  inches. 

8.  A  \yal1,  which  was  to  be  built  24  feet  high,  was  raised  8  feet 
by  6  men,  in  12  days  :  llow  many  men  must  be  employed  to  finish 
the  wall  in  four  tlays  ? 

n.     ft.  m. 

And, 


ft.  m.     ft. 

m. 

As  8  :  q  ::  24—8  : 
d.     m. 
As  12  :  12  ; 

12  to  finish  it  in  12  days. 
d.     m. 
::  4  :  3G  to  finish  in  4  days. 

9.  There  is  a  cistern  having  a  pipe,  which  will  empty  it  in  6 
hours  :  How  many  pipes  of  th^  same  capacity  will  empty  it  in  20 
minutes  ? 

h.  pi.    mi.    pi. 
As  G  :  1  ::  20  :  18  Ans. 

10.  If  a  field  will  feed  G  cows  91  days,  hovy  long  will  it  feed  21 
cows  ?  Ans.  26  days. 

1 1.  How  much  in  length,  that  is  13|  poles  in  breadth,  will  make 
a  square  acre  ?  Ans.  lly^J*- poles. 

12.  If  a  suit  of  clothes  can  be  made  of4i  yards  of  cloth,  1|  yard 
wide  ;  how  many  yards  of  coating-^  of  a  yard  wide,  will  it  require 
for  the  same  person  ?  Ans.  6yds.  Iqr.  340. 

Abbreviations. 

To  kno-^  rvhether  a  fraction,  when  abbreviated^  be  equivalent  in  all 
respects  to  the  07'iginal  fraction. 

Rule. 

As  the  numerator  qi'  the  fraction,  in  its  lowest  terms,  is  to  its  de- 
nominator ;  so  will  the  numerator  of  the  original  fraction  be  to  its 
own  denominator. 

Or,  as  one  numerator  is  to  the  other  ;  so  will  one  denominator 
be  to  the  other,  &c. 

A  owes  B  £75  13s.  6d. ;  now  £100  of  A's  money  is  equa!to£l40 
of  B's  ;  what  must  A  pay  to  satisfy  the  said  debt? 


I 


148  DOUBLE  RULE  OF  THREE, 

£     s.  d. 

]-U=h  therefore,  75  13  6 

5 


7)378     7  G 

£54     1  Of  Ans. 
Now,  to  prove  whether  ^  be  equal  |||. 

Num.  Den.  Num.  Den.  Num.  Num.  Den.  Den. 

As  5  :   7  ::    100  :  140  Or,  as  5  :    100  ::   7   :    140. 

All  questions  in  the  Rule  of  Three  Direct  or  Inverse,  may  be 
wrought  by  the  following 

Rule. 
State  the  questions  as  directed  in  the  Rule  of  Three  Direct ; 
then  multiply  the  second  term  by  the  first  or  third  term  according- 
ly as  the  answer  ought  to  be  greater  or  less ;  divide  the  product 
by  the  other  term,  and  the  quotient  will  be  the  answer. 


COMPOUND  PROPORTION, 

OR  DOUBLE  RULE  OF  THREE, 

TEACHES  to  resolve  such  questions  as  require  two  or  more 
statements  by  Single  Proportion,  and  hence  its  name.  There  is 
always  an  odd  number  of  terms  given,  as  five,  seven,  &c.  All 
questions  in  Compound  Proportion  may  be  stated  and  wrought  by 
the  following 

General  Rule.* 

1.  Place  that  term,  which  is  of  the  same  kind  or  quality  with 
the  answer  sought,  for  the  second  term. 

2.  Then,  of  the  two  terms  in  the  question  of  the  same  kind, 
place  the  greater  or  less  on  the  right  for  the  third  term,  and  the 
other  on  the  left  for  the  first  term,  according  to  the  directions  un- 
der the  General  Rule  for  Simple  Proportion.  Arrange  the  two 
remaining  terms  under  the  first  and  third,  on  the  same  principle. 

3.  Find  the  fourth  term  from  the  first  statement,  and  place  it  for 
the  second  term  in  the  second  part  of  the  statement,  and  find  thp 
fourth  term  from  this  statement,  and  it  will  be  the  answer  required- 

Note.  If  there  be  more  than  five  terms  in  the  question,  the  sam^i 
mode  of  statement  must  be  continued,  and  a  third  proportion  form- 
ed, and  so  on,  and  the  fourth  term  found  from  the  last  statement, 
will  be  the  answer  as  before. 

*  This  nile  is  evident  from  the  General  Rule  of  Three,  for  each  statement 
is  a  particular  statement  under  that  Rule.  If,  then,  all  the  separate  dividend?* 
be  collected  into  one  dividend,  and  all  the  divisors  into  one  divisor,  their  quo- 
tient must  be  the  answer.     Thus,  in  Ex.  1 . 

D.   Int.    D.         Int.  M.       Int.        M. 

400X6  400X6  400x6x9 

As  100  :  0  ::  400  :  -j^,  and  as  12  :-^y  -  ::  9  :  l^]o^,-=$lB  Int.  Aus. 


DOUBLE  RULE  OF  THREE.         149 

Examples. 
1.  If  a  principal  of  glOO,  gain  $6  interest  in  one  year,  what  will 
^400  gain  in  9  months. 

Statement  and  Operation. 
D.     Int.       D.  M.  Int.     M. 

As  100  :  6  ::  400     Or,  12   :  6  ::  9 
M.  M.  D. 

12  :        ::   9  100  :        ::  400. 

400X6  400x6 

Then  100  :  6  ::  400  :     ^^^  =24.    And  12  :  -J^ :  9  :  18  Int.  Ans. 

Or  12  :  24  :  9  :  18  Ans. 
6X9  6X9  6X9X400 

Then  12  :6::9  :  —  =4^  and  100  : -j^  ^  400  : -j^^- =  1 8  Ans. 

Or  100  :  4^  :  400  :  18  Int.  Ans. 
In  this  question,  the  answer  sought  is  interest,  and  therefore  $(> 
must  be  the  second  term.  As  ^400  will  gain  more  interest  in  the 
same  lime  than  glOO,  ^^400  must  be  placed  on  the  right  for  the 
third  term,  and  glOO  on  the  left  for  the  first  term.  And  as  the 
sam>2  sum  will  gain  more  interest  in  12  months  than  in  9  months, 
the  9  must  be  placed  under  the  third  term,  and  the  12  under  the 
first  term. 

The  operation  is  obvious  on  inspecting  it. 
Note.  Instead  of  working  two  proportions,  the  whole  may  be 
reduced  to  one,  by  multiplying  the  Jirst  terms  together,  and  also 
the  third  terms,  and  using  their  products  for  the  first  and  third 
terms.     This  is  merely  changing  the  order  of  the  operation,  as 
*-wilI  be  seen  in  the  preceding  example. 
D.  Int.       D. 
100  :  6  ::  400?  ^,..    ,  . ,     ,, 

12  •      ••    9    (  becomes,  evidently, 

400X0X6 
100X12  :  6  ::  400x9  :     ^^^^^ -=18,  as  before. 

The  work  may  also  frequently  be  contracted  by  dividing  thes 
f.rst  and  third  terms  by  a  common  divisor,  or  the  ^;^r5/  and  second 
terms,  and  using  their  quotients,  for  the  divisor  will  diminish  thfe 
terms  in  the  same  ratio,  and  the  proportion  be  still  preserved. 
'J  hus,  in  the  preceding  example, 

100  :  6  ::  400  becomes  1  :  6  ::  4,  by  dividing  by  100. 
12  :      ::  9  4  :      ::  3,  3. 

And  .  I      ■'  „  [  becomes  1  :  6  :  3  :  18,  Ans,  as  before. 

Ex.  2.  If  950  soldiers  consume  350  quarters  of  wheat  in  7  months, 
how  many  soldiers  will  consume  1464  quarters  in  1  month? 

Ans.  27816  soldiers. 
Ex.  3.  If  1464  quarters  of  wheat  be  used  by  27816  soldiers  in  a 
month,  in  what  time  will  950  soldiers  consume  350  quarters  ? 

Ans.  7  months. 
Ex.  4.  If  144  men,  in  6  days  of  12  hours  each,  dig  a  trench  200 
feet  long,  3  wide  and  2  deep,  how  many  hours  long  is  the  day, 


iov  I>OUBLE  KULE  OF  THREE. 

when  30  men  dig  a  trench  360  feet  long,  6  wide  and  3  deep,  in 
259-2  days  ?  Ans.  7  hours. 

The  following  Rule  for  the  Double  Rule  of  Three,  involves  the 
consideration  of  Direct  and  Inverse  Proportion.  Though  the 
General  Rule  will  enable  the  student  to  solve  all  questions  with 
ease,  this  Rule  is  retained  for  the  satisfoclion  of  those  who  might 
desire  to  use  it. 

RULK. 

Always  place  the  three  conditional  terms  in  this  order  :  That 
number,  which  is  the  principal  cause  of  gain,  loss  or  action,  pos- 
sesses the  first  place  ;  that,  which  denotes  the  space  of  lime,  dis- 
tance of  place,  rate,  medium  or  mean  of  action,  the  second  ;  and 
that,  which  is  the  gain,  loss  or  action,  the  third :  This  being  done, 
place  the  other  two  terms  which  move  the  question,  under  those 
of  the  same  name,  and  if  the  blank  place,  or  term  sought,  fall 
tjnder  the  third  place,  then  the  question  is  in  direct  proportion : 
therefore, 

Rule  I  * 

Bltiltiply  the  three  last  terms  together,  for  a  dividend,  and  the 
two  first  for  a  divisor : — But,  if  the  blank  fall  under  the  first  or 
second  place  ;  then,  the  proportion  is  inverse  ;  therefore, 

Rule  II. 
Multiply  the  first,  second  and  last  terms  together  for  a  dividend, 
and  the  other  two  for  a  divisor,  and  the  quotient  will  be  the  an- 
swer. 

Examples. 

1.  If  g  100  gam  $6  in  a  year  ;  vvhat  will  §400  gain  in  9  months  ? 
D.  P.  Mo.  D.  Int. 

100  :   12  ::  6  Terms  in  the  supposition,  or  conditional  terms. 
400  :      9  Terms  which  move  the  question. 

Mere,  the  blank  falling  under  the  third  place,  the  question  is  in 
direct  proportion,  and  the  answer  must  be  found  by  the  first  Rule  ; 
therefore, 

400X  9x6—21600  For  the  dividend,  and, 
i 00X1 2       =1200  For  the  divisor. 

■•  1.  When  the  blank  falls  under  the  tl^ini  term  by  this  mode  of  stalciner.i. 
>l  in  obvious  on  injpcctinjr  the  statement  that  the  proportion  is  direct^  and  tlio 
same  terms  are  taken  to  form  the  dividend  and  divisor  as  in  the  preceding  rule, 
or,  by  two  statements  in  the  Sing^le  Rule  of  Three  direct. 

2.  But  in  Example  2nd,  and  when  the  blank  falls  under  the  first  or  second 
term,  the  proportion  is  inverse.  In  thip  Example,  7nore  principal  and  interest 
require  /^S5  "time,  and,  every  statement  according  to  the  rule  will  make  more 
req\are  less.  The  operation  by  the  rule  is  the  same  as  from  two  statements  by 
the  Single  Rule  of  Three  inverse.  These  statements  on  this  Example  would 
he  iliua 

100X12 
100  ;  12  ::  400  :  "-~ 

100X12         100XJ2X"10 
^•fl :  — -rr  ■•  6  :  — rr,;rr77 — "-?-  {*b3  illustrate?  the  rule. 


4'X)  i-COxO 


m 


DOUBLE  RULE  OF  THREE.         I5i 


D.Pr. 

100: 

400  : 

9 

See 

Mo. 

12  ; 

9 

the  work  at  larsre. 
D.  Int 
::  ^> 

100 
12 

3600 
6 

12j00)216|00( 
12 

18D. 

Ans. 

9G 
96 

2.  If  glOO  will  gain  $0  in  a  3rear ;  in  what  time  will  J5400  gam 
^18?  D.    Mo.      D. 

100  :  12  ::    6  Terms  in  the  supposition. 
400  :        ::  18  Terms  which  move  the  question. 
Here,  the  blank  falling  under  the  2d  place,  the  question  is  in 
reciprocal  or  inverse  Proportion,  and  the  answer  must  be  sought 
by  the  second  rule  ;  therefore, 

100x12x18=21600  For  the  dividend, 
400X  6        =  2400  For  the  divisor. 
D.Pr.  Mo.  D.Int. 
100:  12  ::    6 
400  :        ::  18 
6  12 

2400  21G 

100 


24|00)216|00(9  months,  Ans. 
216 

^.    What  principal,  at  6  per        4.     If  g400   gain   $\S   in    9 
cent,  per  ann.  will  gain  ^18  in    months  ;    what  is  the   rate  per 

9  months  ?  cent,  per  annum  ? 

Pr.     Mo.    Int.  Pr.  Mo.  Int. 

100  :  12  ::    6  400  :    9  ::  18 

9  ::  18  100  :  12  ::  $6  Ans. 
12 

9  216 
6  100 
D. 


64)21600(400  Ans. 
216 


00 


\n^ 


COMPARISON  OF  WEIGHTS,  Lc. 


Here,  the  blank  falling  under  the  first  place,  the  proportion  i* 
inverse,  and  the  answer  found  by  the  second  rule,  as  in  the  last 
example. 

5.  If  8  men  spend  £32  in  13  weeks  ;  what  will  24  men  spend  in 
52  weeks?  Ans,  £384. 

6.  If  the  freight  of  9hhds.  of  sugar,  each  weighing  12cwt.  20 
leagues,  cost  ^50;  what  must  be  paid  for  the  freight  of  50  tierces 
ditto,  each  weighing  2|cwt.  100  leagues  ?      Ans.  ^289  35c.  Iffm, 

7.  There  was  a  certain  edifice  completed  in  a  year  by  20  work- 
men ;  but  the  same  being  demolished,  it  is  necessary  that  just  such 
an  one  should  be  built  in  5  months.  I  demand  the  number  of  men 
to  be  employed  about  it?  Ans.  48  men. 

8.  If  6  men  build  a  wall  20  feet  long,  6  feet  high  and  4  feet 
thick,  in  16  days,  in  what  time  will  24  men  build  one  200  feet  long, 
n  feet  high,  and  6  feet  thick  ?  Ans.  80  days. 

COMPARISOJV  OF  WEIGHTS  AJVD  MEASURES. 

Examples. 
1.  If  78  pence  Massachusetts  be  worth  1  French  crown,  how 
niiany  Massachusetts  pence  are  worth  320  French  crowns  ? 
F.  cr.  d.    F  cr. 
As   1   :  78  ::  320 
78 

2560 
2240 


24960  Ans. 

2.  If  24  yards  at  Boston  make  16  ells  at  Paris,  how  many  ells 
at  Paris  will  make  128  yards  at  Boston  ? 

Bost.         Par.         Bost.         Par. 
As  24yds.  :  16ells  ::  128yds.  :  85iells,  Ans. 

3.  If  60lfe  at  Boston  make  561fe  at  Amsterdam,  how  many  pounds 
at  Boston  will  be  equal  to  350  at  Amsterdam  ? 

Ans.  375ife  Boston. 

4.  If  95ife  Flemish  make  1001b  American,  how  many  American 
pounds  are  equal  to  550fe  Flemish  ?         Ans.  578  J  fife  American. 


CONJOINED  PROPORTION, 

IS  when  the  coins,  weights  or  measures  of  several  countries  are 
compared  in  the  same  question  ;  or,  in  other  words,  it  is  joining 
many  proportions  together,  and  by  the  relation,  which  several  an- 
tecedents have  to  their  consequents,  the  proportion  between  the 
first  antecedent  and  the  last  consequent  is  discovered,  as  well  as 
the  proportion  between  the  others  ia  their  several  respects. 

I 


CONJOINED  PPxOPORTfON.  153 

This  rule  may  generally  be  so  abridged  by  cancelling  equal 
quantities  on  both  sides,  and  abbreviating  commensurables^  that 
the  whole  operation  may  be  performed  with  very  little  trouble, 
and  it  may  be  proved  by  as  many  statings  in  the  Single  Rule  ot" 
Three,  as  the  nature  of  the  question  may  require. 

CASE  I. 

When  it  is  required  to  find  how  many  of  the  first  sort  of  coin, 
weight,  or  measure, mentioned  in  the  question,  are  equal  to  a  given 
quantity  of  the  last. 

Rule. 

Place  the  numbers  alternately,  that  is,  the  antecedents  al  the 
left  hand,  and  the  consequents  at  the  right,  and  let  the  last  num- 
ber stand  on  the  left  hand  ;  then  multiply  the  left  hand  column 
continually  for  a  dividend,  and  the  right  hand  for  a  divisor,  and  the 
quotient  will  be  the  answer. 

Examples. 

1.  Suppose  100  yards  of  America=100  yards  of  England,  and 
100  yards  of  England=50  canes  of  Thoulouse,  and  100  canes  of 
Thoulouse  =  160  ells  of  Geneva,  and  100  eJls  of  Geneva=200  ells 
of  Hamburgh  :  How  many  yards  of  America  are  equal  to  379  ells 
of  Hamburgh  ? 

Antecedents.  Consequents.  Abriged. 

100  of  America       =   100  of  England.  Ant.     Con. 

100  of  England        =     50  of  Thoulouse.  5         8 

100  of  Thoulouse    =   160  of  Geneva.  379 

J 00  of  Geneva         =  200  of  Hamburgh. 
.379  of  Hamburgh  ? 

370X5 
Therefore,       »    =23G|yds.  of  America=379  ells  of  Hamburgh, 

Illustration. 

The  two  100s  of  both  sides  cancel  each  other.  Let  the  last  cy- 
phers  of  the  next  three  antecedents  and  consequents  be  cancelled, 
which  is  dividing  by  10.  Then  divide  the  second  antecedent  and 
consequent  by  5,  and  the  quotients  will  be  2  on  the  side  of  the 
antecedents,  and  1  on  the  side  of  the  consequents  ;  then  2  will 
measure  the  third  antecedent  and  consequent,  and  the  quotients 
will  be  5  and  8.  10  will  measure  the  4th  antecedent  and  conse- 
quent, and  the  quotients  will  be  1  and  2.  Now,  there  being  2  left 
on  each  side,  they  cancel  each  other,  and  as  there  is  no  farther 
room  for  abridging  by  reason  of  the  odd  number  379,  the  opera- 
tion is  finished,  nnd  the  answer  found,  as  before. 

2.  If  20tfe  at  Boston  make  23ife  at  Antwerp,  and  155  at  Antwerp 
make  180  at  Leghorn  :  How  many  at  Boston  are  equal  to  144  art 
Leghorn?  Ans.  107iflfe. 

3.  If  121fe  at  Boston  make  lOib  at  Amsterdam,  lOlb  at  Amsterdam 
12Jfe  at  Paris  :  How  many  pounds  at  Boston  are  equal  to  80ife  at 
Paris  ?  Ans.  80ft3,. 


I 


lo4  ARlirTKATlON  OF  EXCHANGES, 

4.  If  140  braces  at  Venice  be  equal  to  150  braces  at  Legborn, 
and  7  braces  at  Leghorn  be  equal  lo  4  American  yards  :  How 
many  Venetian  braces  are  eqtial  to  32  American  yards  ? 

Ans.  52~^*j. 

5.  If  40ife  at  Newburyport  make  36  at  Amsterdam,  and  90fe  at 
Amsterdam  make  116  at  Dat>rzick  :   How  many  pounds  at  Newbn 
ryport  are  equal  to  260ife  at  Dantzick  ?  Ans.  2242^- 

CASE  II. 

When  it  is  required  to  tind  bow  many  of  the  last  Sort  of  coin, 
weight  or  measure,  mentioned  in  the  question,  are  equal  lo  a  given 
quantity  of  the  first. 

KULK. 

Place  the  numbers  alternately,  begihning  at  the  left  band,  and 
let  the  last  number  stand  on  the  right  hand  ;  then  multiply  the  first 
row  for  a  divisor,  and  the  second  for  a  dividend. 

Examples. 

1.  Suppose  100  yard^  of  America--=100  yards  of  England,  and 
100  yards  of  England-=50  canes  of  Thoulouse,  and  100  canes  of 
Thoulouse=^160  ells  of  Geneva,  and  100  ells  of  Geneva=200  e\U 
of  Hamburgh.:  How  many  eli.i  of  Hamburgh  are  equal  lo  236" 
yards  of  America  ? 


Ant. 

Con. 

Abridged, 

100  Amer. 

==   100 

Eng. 

Ant.         Con. 

100  Eng. 

=     50 

Thoul. 

6             0 

100  Thoul. 

=   160 

Gen. 

236| 

100  Gen. 

==  200 

Hamb. 

236 

I-X  8 

=  379  Ham. 

2361 

Amer. 

5 

Ans. 

This  needs  no  further  illustration.  The  learner  will  readily 
see,  that  this  case  being  the  reverse  of  the  former,  they  are 
proofs  to  each  other. 

2.  If  20fe  at  Boston  make  23115  at  Antwerp,  and  155  at  Antwerp 
make  180  at  Leghorn  :  How  many  at  Leghorn  are  equal  to  144  at 
Boston  ?  An-s.  144^. 

3.  If  12ife  at  Boston  m?ike  lOife  at  Amsterdam,  and  lOOife  at  Am- 
sterdam 120f6  at  Paris:  How  «»any  at  Paris  are  equal  to  oOlfe  at 
Boston  ?  Ans.   80lfe. 

4.  If  140  braces  at  Venice  be  equal  to  150  braces  at  Leghorn, 
and  7  braces  at  Leghorn  he  equal  to  4  American  yards  :  How  many 
American  yards  are  equalto  52,-^^  Venetian  braces? 

Ans.  32  yards. 
,     5.   If  40'lfe  at  Newburyport  make  36  at  Amsterdam,  and  OOfe  at 
Amsterdam  make  116  at  Drmtzick  :    How  many  pounds  at  Oant- 
zick  are  equal  to  244  at  Newburyport  ?  Ans  283^|lfe. 

JIRBITRATION  OF  EXCHANGES. 

By  this  term  is  understood  how  to  choose,  or  determine  the  best 
way  of  remHting  money  from  abroad  with  advantage;  which  i? 
performed  by  conjoined  proportion  :  'i'hus, 


FELLOWSHIP.  Voo 

1.  Suppose  a  merchant  lias  effects  at  Amsterdam  lo  tlie  amount 
org3530,  which  he  can  remit  hj  way  of  Lisbon  at  840  rees  per 
tlollar,  and  thence  to  Boston,  at  8s.  Id.  per  milree  (or  1000  rees  ;) 
Or,  by  way  ofNantz,  at  5|  livres  per  dollar,  and  thence  to  Boston 
at  6s.  8d.  per  crown  ;  It  is  required  to  arbitrate  these  cxchangci^, 
that  is,  to  choose  that  which  is  most  advantageous  ? 

1  dollar  at  Amsterdam  =  840  rees  at  Lisbon. 
1000  rees  at  Lisbon       ==     97d.  at  Boston. 

3530  dollars  at  Amsterdam. 
840x97x3530 

rrTTTTT-; =£1198  8s.  8yV'-  by  wav  of  Lisbon. 

1000x1 
1  dollar  at  Amsterdam  =  5|  livres  at  Nantz. 
6  livres  at  Nantz  =  80  pence  at  Boston. 

3530  dollars  at  Amsterdam. 
5|x  80x3530 
]v6 =ii^l059  by  way  of  Nantz. 

Here  it  may  be  observed  that  tlie  difference  is  J£139  8s.  8y*-„d. 
in  favour  of  remitting  by  way  of  Lisbon  rather  than  by  Nanlz, 
which  depends  on  the  course  of  exchange,  at  that  lime  ;  but  the 
course  may  vary  so,  that^  in  a  short  lime  by  way  of  Nantz  may 
be  better;  hence  appears  the  necessity  and  advantage  of  an  ex- 
tensive correspondence,  to  acquire  a  thorough  knowledge  in  the 
courses  of  exchange,  to  make  this  kind  of  remittance.  ? 

2.  A  merchant  in  England  can  draw  directly  for  1000  piastres  ih 
Leghorn  at  50d.  sterling  per  piastre  ;  but  he  chooses  to  remit  the 
sum  to  Cadiz  at  19  piastres  for  7000  maravedies ;  thence  to  Am- 
sterdam at  189d.  Flemish  for  680  maravedies;  and  thence  to  Liv- 
erpool at  9d.  Flemish  for  5d.  sterling  :  what  is  gained  by  this  cir- 
•;u!ar  remittance,  and  vyhat  is  the  value  of  a  piastre  to  him  ? 

Ans.  Gain  £28   |49.  sterling  nearly. 
Value  of  a  piastre  5Gd.  3-55qr.  sterling. 

3.  A  merchant  in  New  York  orders  JC500  sterling,  doe  him  nt 
London  at  54d.  sterling  per  dollar,  to  be  sent  by  the  following  cir- 
cuit; to  Hamburgh  at  15  niarks  banco  per  pound  sterling  ;  thence 
to  Copenhagen  at  100  marks  banco  for  33  rix  dollars  ;  thence  to 
Bourdeaux  at  one  rix  dollar  for  G  francs  ;  thence  to  Lisbon  at  125 
francs  for  18  milrees  ;  and  thence  to  New  York  at  ^1^  per  milree  : 
did  he  gain  or  lose  by  this  circular  remittance,  and  what  was  the 
arbitrated  value  of  a  dollar  by  this  remittance  ? 

Ans.  He  gained. 
Value  of  a  dollar  was  69d.  sterling  nearly. 


FELLOWSHIP. 

THE  Rules  of  Fellowship  are  those  by  which  the  accompts  of 
:^cveral  merchants  or  other  persons,  trading  in  partnership,  are  so 
adjusted,  ths^t  each  may  have  his  share  of  the  gain,  or  sustain  his 


156  SINGLE  FELLOWSHIP. 

share  of  the  loss,  in  proportion  to  his  share  of  the  joint  stock,  to- 
gether with  the  time  of  its  continuance  in  trade. 

SIJVGLE  FELLOWSHIP 
Is,  when  the  stocks  are  employed  for  any  certain  equal  time. 

As  the  whole  stock  is  to  the  whole  gain  or  loss,  so  is  each  man's 
particular  stock  to  his  particular  share  of  the  gain,  or  loss. 

Proof.  Add  all  the  particular  shares  of  the  gain  or  loss  together, 
and,  if  it  be  right,  the  sum  will  be  equal  to  the  whole  gain  or  loss. 

Examples. 

1.  Divide  the  number  360  jnto  four  parts,  which  shall  be  to 
each  other,  as  3,  4,  6  and  6. 


As  3-{-4-f-6-l-6  :  3G0  :: 


Answer. 


360  Proof 


2.  A,  B,  C,  and  D  companied  ;  A  put  in  £145;  B,  £219;  C, 
£378,  and  D,£417,  with  which  they  gained  £569  :  What  was  the 
share  of  each  ?  £     s.  d. 

Whole  stock  Gain     C  ^^^  *     '^^    3  8^  |H-|  A's  share. 

A«14^  •  91Q  LAftZAi7. VfiQ   .  )219  :  107  10  31  J^  B's  ditto. 
Asl45-r219-ho78+417:569::  ^3^3  .  ^^^  ^j   g-   ^  C's  ditto. 

(417  :  204  14  51  Tj'j%  D's  ditto. 

£569  —  Proof. 

3.  A,  B,  C,  and  D  are  concerned  in  a  joint  stock  of  ^1000;  of 
which  A's  part  is  §150  ;  B's  $250  ;  C's  g276,  and  D's  ^325.  Up- 
on the  adjustment  of  their  accompts,  they  have  lost  ^337  50^. 
What  is  the  loss  of  each  ?  Ans.  A's  loss  ^50  62Ac.  B's  g84 
37J-C.  C's  $92  811c.  and  D's  gl09  685c. 

4.  A  and  B  companied  ;  A  out  in  £45,  and  took  |  of  the  gain  ; 
"What  did  B  put  in  ?  5—3—2.     Thon,  As  3  :  45  ::  2  :  30  Ans. 

6.  A,  B  and  C  freighted  a  ship  with  68900  feet  of  boards  ;  A  put 
in  16520  feet;  B  28750;  and  C  the  rest;  but  iqa  storm,  the  cap- 
tain threw  overboard  26450  feet  :  How  much  must  each  sustain  of 
the  loss?  ,\ns.  A,  6341f  feet.     P,  n036f  and  C,  90711  do. 

6.  A  gentleman  died,  leavir)g  three  sons  and  a  daughter,  to  whom 
he  bequeathed  his  estate  in  the  following  manner  :  To  the  eldest 
son,  he  gave  312  moidores,  to  the  second,  312  guineas,  to  the  third, 

*  That  thoir  gain  oi-  loss,  in  this  rule,  is  in  proportion  to  their  stocks  is  .evi- 
dent :  For,  as  the  times,  in  which  the  stocks  are  in  trade,  aio  equal,  il" I  put  in  I 
of  the  whole  stock,  I  ou^ht  to  have  -i  of  the  g^ain  :  li  my  part  of  the  stock  be 
j,  my  share  oi  the  gain  or  loss  ought  to  he  i  also.  And  generally  the  same  ra- 
tio that  the  whole  stock  has  to  the  whole  fi:ain  or  loss,  must  e^h  person's  pav- 
:  f icular  stock  have  to  iiis  respective  gain  or  loss. 


SINGLE  FELLOWSHIP.      '  157 

312  pistoles,  and  to  the  daughter,  312  dollars  ;  but  when  his  debts 
were  paid,  there  were  but  312  half  joes  left:  What  mu!*t  each 
have  in  proportion  to  the  legacies  which  had  been  bequeathed  them? 
Ans.  1st  son  £293  Os.  3d.— 2d.  son  £227  17s.  lOfd.— 3d.  son 
X  179  Is.  2id.  and  the  daughter  £48  16s.  8id. 

7.  A  ship,  worth  ^3000,  being  lost  at  sea,  of  v\hich  }  belonged 
to  A,  i  to  B,  and  the  rest  to  C  :  What  loss  will  each  sustain,  sup- 
posing ^450  to  have  been  insured  upon  her? 

Ans   A's  loss  $312  50c. 
B's  937  50 

C*s  625 

8.  A  and  B  venturing  equal  sums  of  money,  cleared  by  joint 
trade  ^140:  By  agreement,  as  A  executed  the  business,  he  was 
to  have  8  per  cent,  and  B  was  to  have  5  per  cent. :  What  was  A 
allowed  for  his  trouble  ? 

^      (fi         (3  (3  <S  ^       d         ^  ^  (S. 

ff)     iP        IP  iP         iP  iP      ui        P  iP         ir> 

As  84-5  :  140  ::  8  :  86^2-  And,  as  8+5  :  140  ::  5  :  53if 

Ans.  $32  30c.  ^j^nu 

9.  A  bankrupt  is  indebted  to  A  £120,  to  B  £230,  to  C  £340. 
and  to  D£450,  and  his  whole  estate  amounts  only  to  £560  :  How 
must  it  be  divided  among  the  creditors  ? 

Ans.  A,  £58  18s.  ll^d.  B,£ll2  19s.7fd.  C,£167  Os.  4d.  and  D. 
£221  Is.  Oid. 

10.  A,  B,  and  C  put  their  money  into  a  joint  stock  ;  A  put  in  $40  ; 
B  and  C  together  $170  :  They  gained  $126,  of  which  B  took,g42 : 
What  did  A  and  G  gain,  and  B  and  C  put  in  respectively  ? 

Ans.  $24  A's  gain,  $70  B's  stock,  $100  C's  stock.  $60  C's  gain 

11.  A,  B,  and  C  companiod  ;  A  put  in  £40  ;  B  60,  and  C  a  sum 
unknown :  They  gained  £72  ;  of  which  C  took  £32  for  his  share  ; 
What  did  A  and  B  g:ain,  and  C  put  in  ? 

Ans.  £16  A's  gain,  £24  B's  gain,  and  £80  C's  stock. 

12.  A,  B,  and  C  put  in  $720,  and  gained  $540,  of  which,  so  oft- 
ten  as  A  took  up  $3,  B  took  5,  and  C  7  :  What  did  each  put  in  and 
gain  ? 

Instead  of  the  above  rule,  you  may  find  a  common  multiplier  to 
multiply  the  proportions  by,  or  multiplicand  to  be  multiplied  by  the 
given  proportions,  thus,  15)720(48  multiplicand  to  find  the  stock?. 
And  15)540(36  multiplicand  to  find  the  gains. 

^  '  $ 

48x3=144  A's  stock,  i  C  36x3=108  A's  jrain. 

48X5=240  B's  ditto.   >  And   ?  36x5=180  B's  ditto. 
48X7=336  C's  ditto.  )  (  36x7=252  C's  ditto,  as  before. 

13.  A,  B,  C,  and  D  companied  ;  and  gnin.  d  a  sum  of  monry  of 
which  A,  B  and  C  took£i20,  B.  C  and  D,£l80,  C,  D  and  A,  £160, 
^nd  D,  A  and  B,£140  :   What  dij>tinrt  gain   had  e xrh  ? 

The  sum  of  the^e  4  num'ifrs  i.s£6G0.  and  as  each  man's  money 

is   named   3    ti.res.   th'^refjue  |^  viz.  £200  is  \hr    whole  gain 

Therefore  £200— £  120  A'fi  B'-'  and  C's  ^air.  =  ]e80  D's  gain  ;— 
And  £200— .£  180  B's,  C*s  and  D'5  -air.=  £20  A'x  gsin.--£ '^OO— 
£160  C's.  D's,  and  A's  gain=£40  B's  gain.— And £200— £  140 
D's,  A's  and  B's  gain=£60  C's  gain. 


\b^  SINGLE  FELLOWSHIP. 

14.  Two  merchants  companied  ;  A  put  in  £40,  and  B  288  do- 
cats.  They  gained  £135,  of  which  A  took  £60.  What  was  the 
value  of  a  ducat  ? 

As  £60,  A's  gain  :  £40,  his  stock  ::  £135  the  whole  gain— £  60, 
A's  gain  :  £50,  B's  stock. 

Due.    £     Due.  s.  d. 
And,  as  288  :  60  ::  1  :  3  5|  Ans. 

15.  Four  men  spent,  a^a  reckoning,  20  shillings,  of  which  they 
agreed  that  A  should  pay  |,  B,  ^,  C,  |,  and  D,  |.  What  must  each 
liay  in  that  proportion  ? 

s.      d. 

9     2-j 
G      1] 


^,  Answer. 


IG.  A,  B,andC  companied  ;  A  put  in  £40-25;  B  £80*5 ;  and  C 
£  161  r  they  gained  £120.     What  is  each  man's  share  ? 

££££££ 
4i;)25+80-5+161  :  120  ::  40-25  ;  17-142475  =  A's 

34-28495  =B's 
G8-5699     =C's 


Proof  £119-997325 

17.  A,  B,  C,  and  D  gain  ^200  in  trade,  of  which  as  often  as  A  has 
^6,  B  must  have  glO,  C  gl4,  and  D  g20  :  What  is  the  share  of 
t'ach  ?  Ans.  A's  share  ^24,  B^s  jg40,  C's  $56,  and  D's  $Q0. 

18.  An  insolvent  estate  of  ^633  60c.  is  indebted  to  A,  ^312  75c. 
to  B,  ^297,  to  C,  $50  25c.toD,g0  25c.toE,g200  toF,gl42  50c. 
and  to  G,  ^^21  25c.  ;  what  proportion  will  each  creditor  receive  ' 

$      c.     ■ 
Ans.     A's  share  =  193  51  41 


B's   - 

-   183  76-^1 

C's   - 

31  09-2:> 

D's  - 

0  15-41 

E's 

-   123  75- 

F's  • 

-   88  17-18 

G*s   - 

13  14-1^7 

Proof  ^633  59-97 

I'?.  A  sHtp  vias  driven  on  ?hore  in  a  gale,  and  in  Iighle?)ing  and 
getting  her  afloat  again  and  in  reloading,  an  expense  of  <;§763  was 
incurred  ;  the  ship  was  valued  at  ^10000,  freight  at  Ji3200,  molas- 
ses owned  by  A,  at  ^5200,  sugar  owned  by  B,  at  j^4700,  and  rum 
oivned  by  C,  at  J2500 :  how  much  is  this  loss  on  every  $100,  and 
Ii?)w  much  must  each  parly  pay  of  it  *? 


DOUBLE  FELLOWSHIP.  15^ 

^  ^        $        $     $ 

0000+3200-1-52004-4700+2500=23600.  As  25600  ;  768  ::  lOO:.^' 
g       g         g         §  Ans.  on  each  $100. 

Then,  As  100  :  3  ::  1000  :  300  to  be  paid  by  the  ship, 
320  :     96         -         -        -freight, 
5200  :  156         -         -  A, 

4700  :  141  .       -         -  B, 

2500  :     75         -         -  C,  Ans. 


768  Proof. 

20.  A  vessel,  valued  at  g  13000  was  laden  with  hardware  for  E 
vahied  at  J3000,  with  cordage  for  F",  at  goOOO,  with  dry  goods  for 
G,  at  ^3200,  with  goods  for  H,  at  J7900,  and  for  \,  at  $4400  ;  the. 
captain  was  obhged  to  prevent  sinking  in  a  storm  to  throw  over- 
board three  fifths  of  the  hardware,  and  two  fitlhs  of  the  cordage, 
with  goods  of  IT  valued  at  $2700;  allowing  the  freight  ti  be  $3500, 
what  will  be  the  average  of  the  loss  on  100  dolls,  and  what  must 
be  paid  to  E,  F,  and  U,  for  their  property  thrown  overboard  ? 

Ans.  $16  25cts.  on  $100,  and  E,  F,  and  II  must  receive  togeth- 
er $5443  75cts. 

Note.  If  the  property  of  E,  F,  and  11,  had  "been  insured,  the  re- 
mainder of  their  loss  must  be  paid  by  the  insurers.  See  Policiej' 
of  Insurance. 

DOUBLE  FELLOWSHIP;'' 

Or,  Fellowship  with  Time,  is  occasioned  by  the  shares  of  part- 
ners being  continued  unequal  times. 

KULF.. 

Multiply  each  man's  stock,  or  share  by  the  time  It  was  continu- 
ed in  trade.     Then, 

As  the  whole  sum  of  the  products,  is  to  the  whole  gain  or  loss, 
%o  is  each  man's  particular  product,  to  his  particular  share  of  the 
gain  or  loss. 

Examples. 
1.  A,  B,  and  C  hold  a  pasture  in  common^  for  which  they  pay 
£40  per  annum.     A  put  in  9  oxen  for  5  weeks  ;  B,  12  oxen  for 
7  weeks,  and  C  8  oxen  for  16  weeks.     What  must  each  pay  of  the 
rent  ? 

0x5=45.     12x7=84,  and  8x16  =  128,  then  1284-844-45=257, 
As  257  :  40  ::  45        As  257  :  40  ::  84  As  257  :  40  ::  128 
45  84  40 

£   s.    d. 

200  160  257)5120:i9  18  5-/A 

160  320 

. £  s.  (1.  £    g.   d. 

257)1800(7  0  0|f»-  257)3360(13  1  5i|| 

*  When  times  are  equal,  the  shares  of  the  gain  or  loss  are  evidently  as  the 
stocks,  as  in  Single  Fellowship  ;  and  when  the  stocks  are  equal,  the  shares  *jr<- 
as  the  times  ;  wliereforc,  when  rjeither  aro  eqnnl.  the  shav-s  nv\-t  hp  a,'  ^h"'' 
products. 


60  DOUBLE  FELLOWSHIP. 


£ 

s. 

(1. 

A's  = 

7 

0 

o|M 

B's  = 

13 

1 

HU 

C's  = 

19 

18 

h%\ 

Proof  40 

0 

0 

2.  Four  merchants  traded  in  company  ;  A  put  in  ^400  tor  i\vv, 
months,  B,  gGOO  for  7  months,  C,  g960  for  8  months,  and  D,  ^1200 
for  9  months ;  but  by  misfortunes  at  sea,  they  lost  ^750.  What 
must  each  man  sustain  of  the  loss. 

Answer      \  ^'  ^^^  ^^"-  ^^>'     ^'  ^^^^  ^^^^  Vo^-  I 
^n:,wer,     ^  g    142  40     6-^V  ^     284  81       Ol|      ^ 

3.  A,  with  a  capital  of  £100  began  trade  January  1st,  1787,  and 
meeting  with  success  in  his  business,  he  took  in  B  as  a  partner,  on 
the  1st  day  of  March  following,  with  a  capital  of  £150.  Three 
months  after  that,  they  admit  C  as  a  third  partner,  who  brought 
into  stock  £180,  and  after  trading  together  until  the  1st  of  Janua- 
ry, 1788,  they  found  there  had  been  gained  since  A's*  commencing 
business  £177  13s.     How  must  this  be  divided  among  the  partners  '.' 

Ans.  A, £53  IGs.  8d      B,£67  5s.  lOd.     C,£56   10s.  6d. 

4.  Two  merchants  entered  into  partnership  for  18  months  ;  A, 
at  first,  put  into  stock  ^400,  and  at  the  end  of  8  months  he  put  in 
^200  more  ;  B,  at  first,  put  in  gllOO.  and  at  4  months'  end  took 
out  §280.  Now  at  the  expiration  of  the  time,  they  found  they  had 
gained  gl052.     What  is  each  man's  just  share  ? 

Ans.  A,  §385  90c.     B,  $66G   10c. 

3.  A  and  B  companied  ;  A  put  in  the  1st  of  January,  £150  ;  but 
B  could  not  put  in  any  until  the  1st  of  May  :  What  did  he  then  put' 
in,  to  have  an  equal  share  with  A  at  the  year's  end  ? 

Ans.  £225. 

6.  E,  F,  and  G  companied  ;  E  put  in,  the  first  of  March,  £30, 
F,  the  first  of  May,  put  in  80  yards  of  broadcloth  ;  and  on  the  1st 
of  June,  G  put  in  §120.  On  the  1st  of  January  following,  they 
reckoned  their  gains,  of  which  E  and  F  took  £228.  F  and  G£215 
10s.  and  G  and  E  £187  10s.  What  was  the  whole  gain,  and  tit 
gain  of  each  ?  What  did  they  value  a  yard  of  cloth  at  ?  and,  what 
was  G's  dollar  worth  ? 

2281.-1-2151.  10s.-f-187l.  103=6311.  and  6311 -f-2=315].  10s.  the 
whole  gain;  then,  3161.  10s.— 228=871.  10s.  G's  gain.  3151.  10s. 
—2151.  10s.  =  1001.  E's  gain,  and  3151.  10s  —1871.  10s.=  1281.  F's 
gain.  To  find  the  value  of  one  yard  of  cloth,  say,  A>*  1001.  E's 
gain  :  301.  his  stock  ::  1281.  F's  gain  :  381.  8s.  ;  then,  inversely. 
As  19  months  :  381.  83.  ::  8  months  :  481.  the  value  of  the  whole 
cloth. 

As  80yds.  :  481,  ::  1yd.  :  12s.  answer.  Now,  to  find  the  value 
of  a  dollar.  As  1001.  E's  gain  :  301.  his  stock  ::  871.  10s.  G's  gain  : 
261.  3s.  ;  then,  inversely,  As  10  months  :  261.  6s.  ::  7  months  :  371. 
10s.  =  120  dollars.  Lastly  r  As  120  dollars  :  371.  10s.  ::  1  dollar  : 
6s.  3d.   Answer. 


PRACTICE.  161 

7.  E,  F  and  G  companied  ;  E  put  in  ^400  for  '75  of  a  year  ;  F 
<J300  for  '5  of  a  year,  and  G  g500  for  -25  of  a  year  ;  with  which 
they  gained  $120  :  Required  the  share  of  each. 

400x-75=300 
300X  -6=150 
.500X-25=125 

.  g 

575  :  720  ::  300  :  375if 
187if 

Proof  720  Dollars. 

8.  A  put  in  i  for  f  of  a  year,  B  |  for  i  a  year,  and  C  the  rest 
for  one  year  ;  their  joint  stock  was  1,  and  their  gain  1  ;  what  ia 
each  share  ?  Ans.  A's  is  ^ 

B's      ^ 
C's      ^ 

Proof  =1 

9.  A  and  B  entered  into  partnership  for  16  months.  A  put  in 
gl200  at  first,  and  9  months  afterwirds  J200  more  ;  B  put  in  at 
first  j^l500,  and  at  the  end  of  6  months  took  out  §500 ;  their  gain 
was  jj772  20c.  ?  what  is  the  share  of  each  ? 

Ans.  A's  share  J^401  70c.     B's  share  §370  50c. 


PRACTICE, 

IS  a  contraction  of  the  rule  of  Three  Direct,  when  the  first 
term  .happens  to  be  a  unit,  or  one  ;  and  has  its  name  from  its 
daily  use  among  merchants  and  tradesmen,  being  an  easy  and  con- 
cise method  of  working  most  questions  which  occur  in  trade  and 
business. 

The  method  of  proof  is  by  the  Rule  of  Three,  Compound  Mul- 
tipHcation,  or  by  varying  the  order  of  them. 

A  variety  of  rules,  adapted  to  particular  cases,  is  usually  given 
under  Practice.  Most  of  the  sums,  however,  fall  under  two  heads, 
and  may  be  wrought  by  two  General  Rules,  adapted  to  these  cases. 
On  account  of  their  great  practical  importance,  these  two  rules 
should  be  thoroughly  understood. 

General  Rule  I. 
When  the  price  of  1  yard,  life,  ^'C.  is  given  to  find  the  value  of  any 
number  of  yards,  4*c. 

1.  Suppose  the  price  of  the  given  quantity  to  be  II.  ID.  Is.  &c. 
then  will  the  quantity  itself  be  the  answer,  at  the  supposed  price. 

2.  Divide  the  given  price  into  aliquot  parts,  either  of  the  sup- 
posed price  or  of  one  another,  and  find  the  quotients  of  the  several 
aliquot  parts  ;  and  their  sum  will  be  the  truje  answer, 

W 


U2 


PRACTICE. 
Example. 


What  Is  the  value  of  468  yards,  at  2s.  O^d.  per  yard  ? 
JS468   s.  d.  Answer  at  £1  s.  d. 


28.  6d.  is  } 

3d.   is   yV 
Ad.   is  yV 

=  58  10  0 
=     5  17  0 
=     099 

ditto  at  0  2  6 
ditto  at  0  0  3 
ditto  at  0  0  0| 

The  full  price 

=  je64  16  9 

0  2  9^ 

In  this  example  it  is  plain,  that  the  quantity  468  is  the  answer  at 
£  1  J  consequently  as  2s.  6d.  is  |  of  a  pound,  i  part  of  that  quanti- 
ty, or  J£68  10s.  is  the  price  at  2s  6d. ;  in  like  manner,  as  3d.  is  the 
■j'^  part  of  2s  6d  so  y^  P^^^  of  ^^  10s.  or  £5  17s.  is  the  answer 
at  3d.  and  as  jd.  is  ^^  of  3d.  so  y'^  of  £5  17s.  or  9s.  9d.  is  the  answer 
at  ^d.  Now,  as  the  sum  of  all  these  parts  is  equal  to  the  whole 
price  (2s.  9Ad.)  so  the  sum  of  the  answers  belonging  to  each  price 
will  be  the  answer  at  the  full  price  required,  and  the  same  will  be 
true  in  any  example  whatever. 

Before  the  questions,  hereafter  given,  can  be  wrought,  the  fol 
lowing  Tables  must  be  perfectly  gotten  by  heart. 


TABLES. 

Aliquot^  or  even  parts  of  Money. 


Pis.  of  a  jhil.  of  a  £.     | 

d. 

s.         £ 

6 

=  i  =  tV 

4 

=  i  =  ^V 

3 

=   i   =    s  V 

2 

=    i    =  tIo 

n 

=   i    =  tIo 

1 

I 

=  ^V  =  ijo 

X 

4 

tV    ~   fl6  0 

Farts  of  2  Shill.         | 

d. 

2s. 

1 

H 

2 
3 

24 

=         tV 
=         tV 

4 
6 

=              JL 
=              1 

Part 

3  of  a  Pound. 

Farts  of 

a  DoHar 

S. 

d. 

£ 

c. 

$ 

10 

0  = 

\ 

60 

=  i 

6 

8  = 

\ 

331 

s" 

6 

0  = 

\ 

25 

=         4 

4 

0  = 

\ 

20 

3 

4  = 

i 

16f 

=        1 

2 

6  = 

i 

121 

=        i 

1 

8   = 

iV 

l^ 

=    tV 

1 

4  == 

1  5 

H 

=    tV 

1 

3  = 

tV 

5 

=  h 

1 

0  = 

io 

4 

=         oV 

0 

10  == 

.V 

^ 

=  ^ 

0 

8  = 

sV 

2 

=  l^i 

0 

5  = 

4V 

1 

=  rh 

0 

4  = 

aV 

0 

3  = 

aV 

0 

2  = 

T20 

PRACTICE. 

Aliquot,  or  even  Paris  of  Weight. 

Parts  of  a  Cwt. 

Parts  of  i  Cwt. 

Parts  of  i  Cwt. 

Parts  of 

Qrs.  Ife         Cwt. 

ife               X  Cwt. 

ife             1  Cwt. 

Cwt.  qr. 

2     0     =     i 

28          =       ^ 

14         =      1 

10  0 

1     0     =     i 

14          =       i 

7          =       i 

3  0 

0  16     =     \ 

8          =       1 

4=1 

4  0 

0  14     =     1 

7         =       i 

2          =      rV 

2  2 

0     8     =    Jj 

4    =  ^,  i          -- 

2  0 

<^     7     =     yV 

1   1 

0     4     =    ^V 

1 

1  0 

larj 


Ton. 


—  1 

—  IT 
=  ^ 
I 

5 

=  i 

=  tV 

=  iV 


Another  Table  of  aliquot  Parts  of  Money. 


Parts  of  a  shill. 
rl. 

10  = 

9         = 
8         = 


s. 


7* 


Parts  of  a  Pound.. 


s.   d. 

18  0 

17  6 

16  8 

16  0 

15  0 

14  0 

13  4 

12  6 

12  0 

8  0 

7  6 

6  0 


=  I 


_8 
1  O 
3 

4 

a_ 

3 

1. 

0 
_«_ 
1  0 

1  0 


Parts  of  a  Dollar. 

C. 

D. 

931 

= 

if 

91| 

= 

H 

90 

= 

A 

871 

:;= 

i 

83i 

=. 

6 

m 

= 

u 

80 

2=S. 

75 

=, 

70 

= 

t'j 

68* 

= 

U 

66f 

=s. 

I 

60 

= 

68J- 

= 

t\ 

56i 

= 

tV 

43| 

= 

tV 

41f 

= 

w 

40 

= 

s 

37A 

= 

1 

3U 

= 

.V 

30 

= 

A 

m 

== 

tV 

^  S.  (1. 

H  per  cent=0  3 

^ =0 

5 =1 

64 


A  TABLE  OF  DISCOUNT  PER  CENT. 


=  1 
=  1 


6 

0 
3 
6 


£  s.  d-l 

81  percent.=l  9 

10  =2  0 

15  

l^i 

20  


=2  6  1 
=3  0, 

=3  6h 

=4  Oj  • 


s.d. 
22Jper  cent.=4  6 


25 
30 
35 
40 
45 
50 


=5  0 

=6  0  I,  ? 
=7  0 
=8  0 
=9  0 
=  10  0 


#• 


164 


PRACTICE. 


Examples. 
1.  What  will  354J  yards  cost,  at  ^d.  per  yard? 

s.    d. 
|{d.(jV|354  6  value  of  354^  yards,  at  Is.  per  yard. 

Ans.  £0  7  4^  value  of  354^  yard,  at  Jd.  per  yard. 

Or  thus.             Or  divide  by  8  and  6,  thus,  8)354  6 
£    s.  d.      s.    d.  


8)17  14  6=354  6 


6)  2     4  3f 


6)  44  3£ 


7  4^  Ans.  as  be  f. 


7  4i  Ans.  as  before. 

2.  What  will  759|  yards  come  to,  at  3d.  per  yard  ? 
3d.)i|759  9  value  at  Is.  per  yard. 

£    s.     d. 

2|0)18|9  Hi         Or  thus,  |3d.|i|37  19     9  value  at  Is.  per  yard. 

Ans. £9  9  11}  value  at  3d.   Ans.   £9     9  111  value  of  759fyds.  at 
per  yard 3d.  per  yard. 

3.  What  is  the  cost  of  227yds.  at  50  cents  per  yard  ? 

^'$  $ 

50l||227=price  at  $1  per  yard. 


^113  60c.  Ans. 
4.  What  cost  927yds.  at  53|  cents  a  yard  ? 

o    i  $ 


50 
31 


927=price  at  $1  per  yard. 


463  60=price  at  50  cents. 
30  90=    do.   at  3-^  cents. 


g494  40c.  Ans. 


Questions, 
yds. 


Answers. 
£ 


6.  918i  at 

6.  739?-  - 

7.  667i  - 

8.  475i  - 

9.  4871  - 

10.  568^  - 

11.  649i  -   lO'd. 


Id.  per  yard    118     33. 


1  d. 
Hd. 

2  &. 
5  d. 

7J-d. 


3 

3 

5 

10 


1     71 
10   1 
14 

3 


Quest. 

yds. 

13    265 

14.   2691  -   16£ 


els. 
at  121 


15.1050 
3|  16.  618 


12.   164 


11 2d. 


17   15 

27      1 

8     0 


o' 

0^ 

r 


7.  328 

18.  817 

19.  296 

20.  300, 


21.  7581yds.  at  Is.  9d.  per  yard, 
d.     £    s.  d 


37  18  6  =  value  at  Is.  per  yard. 
18   19  3  =     do.    at  6d. 
9     9  71=     do.    at  3d. 


£66    7  4^  Aos, 


871 
57i 
30^ 
15 


Ans. 

$     c.  m. 

33   12  5 

44  91  7 

65  62  5 

540  75  0 

188  60  0 

245   10  0 

44  40  0 

52  58  71 


PRACTICE. 


16( 


22. 


106yds 
4s 
8d. 
1 


23. 
24. 
25. 
26. 
27. 


nilbs.  at 


68i 
67i 
614 
1671 


,.  at  4s. 

9id. 

1 

106 

iof4. 

1 

21  :: 

4 

^ 

3  :: 

10 

0  :: 

8 

0  :: 

4 

=  value  at£l  per  yard. 

= value  at  4s. 
8      -     -      8d. 
10      .     -      Id. 
5      -    -     id. 


£25. 


d. 


n 


1! 


:  11 

£ 

3 

2 

27 

41 

491 

163 


2  Ans. 
s.      d. 


10 

5 

11 

1 
4 
6 


8f  Ans. 

81 
3 
0 
3 


Note.  If  there  be  pounds  also  in  the  price  :  Multiply  the  quantity 
by  the  pounds,  and  to  the  product  add  the  value  of  the  quantity 
for  the  other  parts  of  the  price  as  in  the  preceding  examples,  and 
the  sum  will  be  the  answer. 

29.  Find  the  value  of  156yds.  of  cloth  at  £3  6a.  8d.  per  yard. 

£ 

156=value  at  £l  a  yard. 
3 

468=value  at  £3  a  yd. 
52        -        -         6s.  8d.  a  yd. 


69.8d=i 


£520  Ans. 
30.  What  is  the  cost  of  224iyds.  at  £3  7g.  6d.  a  yard  ? 
£  s. 


5s.       =J 
2s.  6d.=i 


224  ::  10 


=cost  at£l  a  yard. 


1122 
56 
28 


10 
2  ::    6d. 
1  ::  3 


£5  a  yd. 
5s.  a  yd. 

2s.  6d.  do. 


£1206  ::  13  ::  9    Ans. 


Answers. 
£     s.     d.  £         s. 

31.  3451yds.  at  6  ::  6  ::  0  per  yard=2159 

32.  59#    -     -    3  ::  6  ::  8      -        -         199 


33.  75 

34.  68 


5  ::  3  ::  4 
4  ::  6  ::  0 


387 
292 


d. 

7  ::  6 
3::  4 

10  ::  0 

8  ::  0 


General  Rule  II.  ^ 

Wlien  the  price  of  one  hundred  weight,  &c.  is  given  of  several  de* 
nominations ^  to  Jind  the  value  of  a  quantity  of  several  denominations 
also. 


im  PRACTICE, 

Multiply  the  price  by  the  integers,  and  take  parts  for  the  rest 
from  the  price  of  an  integer,  and  the  sum  of  the  product  and  quo- 
tients will  be  the  answer. 

Examples. 
1.  If  1  Cwt.  cost  £4  17s.  4d.  what  will  9  Cwt.  3qrs.  14lb.  cost 
at  the  same  rate  ? 

2qrs.  Oft  is  t\    4  17  4=price  of  1  Cwt. 
Iqr.    0         X  9=N umber  of  Cwt. 

0       14 


43  16  0=co9tof  9  Cwt, 

2  8  8=             0  2qrs. 

1  4  4=             0  Iqr. 

0  12  2=;=             0  0  Hlfe 


Ans.  £48     1  2=  Cwt.  9  3  14ife 
Cwt.  qr.  Ife  £    s.    d.  £      s.     tl. 

Ex.2.     8   1   16  Sugar  at  5  17     9  per  Cwt.  =49     8     2| 

3.  7  3  19         -         7   12     8     -         -       CO     9     0^ 

4.  12  1  24  -         3   18   10     -         -       49     2     7-} 

5.  72  3  27         -         8   11     5     -         -     625   U    lOf 

6.  16  2  17   Coffee     2  15  11     -         -       46   11      1 

7.  271fe  lOoz.  0     14        per  ife        1   16  10 

8.  13  10  12pwt.  8grs.  silver  at  £4  7s.  6d.  a  Ife  60   14  11^ 

9.  I7oz.6pwt.  16grs.ofGold£3  16  Sd.peroz.  66  8  lOf 

10.  3  Tun  2hhd.  48gall.  of  wine  at  £5   16  9= 

11.  7  Acres  3  roods  15  rods  16  feet  at £7  lis.  9d.  an  acre=^ 

Though  the  Genera!  Rules,  given  above,  are  sufficient  for  an- 
swering questions  in  Practice,  yet  some  may  perhaps  be  answered 
more  easily  by  other  rules.     Several  cases  fdlow. 

CASE  I. 

J'Vhen  the  price  is  any  even  number  of  shillings  uncUr  24  :  IVTultipIy 
the  given  quantity  by  half  the  price,  and  double  the  first  figure  of 
the  product  for  shillings.     The  rest  of  the  product  will  be  pounds.*- 

N.  B.  If  the  price  be  2s.  you  need  only  double  the  unit  figure 
for  shillings.     The  other  figures  will  be  pounds. 

Examples. 

1st.  What  will  746  yards  cost  at  2s.  per  yard? 
746 


Ans.  £74   12  value  at  23.  per  yard. 

*"  Suppose  the  price  to  be  IGs. ;  as  the  quantity  is  tlie  price  at  £1,  the  aus'.vcr 
will  be  11=.  SL  of  the  quantity,  in  pounds  and  decimals.     The  decimal  is  to  be 

2  0  10  ^  •'  '^ 

doubled  for  its  value  in  shillings,  according  to  a  previous  rule.     The  rule  is  lim- 

;*.ited  to  24s.  because  the  half  of  a  greater  nnnxber,  would  be  an  inconvenient 

multiplier.    The  reasoa.of  the  rules  ia  the  ten  following-  cases  is  sufficiently  ob- 


PRACTICE.  16t 

Note.     The  above  k  done,  by  sayin*  twice  6  (the  unit  figare)  is 
12.     The  other  figures^  viz.  74,  are  pounds. 

2d.  What  will  567|yds.  at  2s.  per  yard  come  to  ? 

Ans.  £56  15s.  6d, 
N.  B.     Before  I  double  the  unit  figure,  viz.  7,  I  consider  thatf 
of  a  yard  at  2s.  per  yard,  will  amount  to  Is  6d.     Then  I  double  7, 
which  makes   14s.  and  Is.  6d.  added,  makes  15«.  6d.     The  other 
figures  are  pounds. 

Questions 
Yds 
3d.     129A  at    4s.   per  yard 

4th.  697    —    6s.    

5th.  845    —    8s.    

6th.   917i  --  10s.     

CASE  II. 

fVhen  the  price  wants  an  even  part  of  2s. :  First  find  the  value  of 
the  whole  quantity  at  2s.  per  pound,  yard,  &c.  then  divide  it  by 
that  even  pari  which  is  wanting,  and  subtract  this  quotient  from  the 
value  at  2s.     The  remainder  will  be  the  answer. 

Examples. 
1.  What  will  95i  yards  cost  at  22d.  per  yard? 

£     s.    d. 
I  2d.  I  xV  1  9     11     0  value  at  2s.  per  yard. 
I         1        I  0     15  11  value  at  2d.  per  yard. 


Answers 

£        s 

d 

25     18 

0 

209       2 

0 

338       0 

0 

458     12 

6 

Ans.  £8     15     1  value  at  Is.  lOd.  per  yard. 
Questions,  Answers. 

yds.  £     s.    d.        • 

2d".     64     at  23d.    per  yd.     6     2     8 

3d.   128    —  22id.  12     0     0 

4th.  2461—  21d      — 2111     4^ 

5th.375|—  20d.     31     5     5"" 

CASE  in. 

When  there  are  pence  in  the  price  which  are  an  even  pari  of  a 
shillings  besides  an  even  number  of  shillings  under  20  :  First  find  the 
value  of  the  quantity  at  the  shillings  per  yard,  &c.  according  to 
Case  1st. :  then  suppose  the  quantity  to  stand  as  shillings  per  yard  ; 
divide  it  by  that  even  party  which  the  pence  are  of  Is.  and  this 
quotient  being  added  to  the  value  before  found,  the  sum  will  be 
the  answer. 

Examples. 
1st.  What  will  156i  yards  come  to,  at  6s.  4d.  per  yard  ? 
Yds. 
1561 

o 


£46   19  0  value  of  1561  yards  at  Cs.  per  yard. 
52s.  2d.  =2  12  2  value  of  ditto  at  4d.  per  yard. 

An^.  £49  11  2  value  of  ditto  at  6s.  4d.  per  yard. 


im  PRACTICE. 

Questions.  AnswerSc 

Yds.        s.  d.  £   8.   d. 

2d.    17^  at     4  OA  per  yd.    3  10  8^. 

3d.    69|  —    6  01  18     2  2|. 

4th.  68i  —    8   1  27   11  8^ 

oth.  96    —  10  n  48   12  0. 

6th.  67J-—  12  2"  41      1   3. 

CASE  IV. 

When  the  price  is  any  odd  number  of  shillings  under  20  :  Find  the 
value  of  jthe  greatest  even  number  contained  in  the  price,  accord- 
ing to  C^se  1st.  and  add  thereto  the  value  of  the  quantity  at  Is. 
per  yard,  kc.  which  sum  will  be  the  answer  :  Or,  Multiply  the 
quantity  by  the  price,  according  to  the  1st  or  2d  Case  in  Simple 
Multiplication,  and  divide  the  product  by  20,  the  quotient  will  be 
the  answer:  Or,  lastly,  if  the  price  be  not  more  than  12s.  find  the 
value  of  the  quantity  at  Is.  per  yard,  kc.  and  multiply  it  by  the 
number  of  shillings  in  the  price  of  1  yard ;  the  product  will  be 
the  answer. 

Examples. 

1st.  What  will  186  yards  cost,  at  3s.  per  yard  ? 
£      s. 

18     12  value  at  2s.  per  yard. 
9       6  ditto  at  Is.  per  yard. 


£27     18  Ans. 


Or  thus. 
£    s. 

9     6  value  at  Is.  per  yard. 
3  i 


Product  £27  18  Ans. 


2d.  What  will  647  yards  cost,  at  17s.  per  yard 
8 


£517  12  value  at  16s.  per  yard. 
32     7  ditto  at  Is.  per  yard. 


Ans.  £549  19  dilto  at  17s.  per  yard. 

Questions.  Answers. 

Yds.         s.                   £     s.  d. 

3d.    169i  at     6  per  vd.    42     6  3 

4th.  248j  —    7 ^     87     1  3 

5th.  139    —    9 62   11  0 

^          ^th.  782    --25 977   JO  0 


PRACTICE. 


IQO 


CASE  V. 

fVlien  the  price  is  shillings,  pence  and  farthings ;  and  not  an  even 
part  of  a  pound  :  Multiply  the  given  quantity  by  the  shillings  in 
ihe  price  of  1  yard,  kc.  and  take  parts  of  parts  from  the  quantity 
for  the  pence^  &c.  then  add  them  together,  and  their  sum  will  be  the 
answer,  in  shillings,  &c.  Or  you  may  let  the  given  quantity  stand  a3 
pounds  per  yard,  &c.  then  draw  a  line  underneath,  and  take  parts  of 
parts  therefrom  ;  which  add  together,  and  their  sum  will  be  the  an- 
swer. 

N.  B.  I  advise  the  learner  to  work  the  following  examples  both 
ways,  by  which  means  he  will  be  able  to  discover  the  most  concise 
method  of  performing  such  questions  in  business  as  may  fall  under 
this  case. 

Examples. 

}.  What  will  248^  yards,  at  7s.  6d.  per  yard,  come  to? 
|6|i|248s.  6d.  value  of  2481  yards,  at  Is.  per  yard. 
7 


1739  6  value  of  ditto  at  7s.  per  yard- 
124  3  value  of  ditto  at  6d.  per  yard. 

2|0)18613  9 

Ans.  £93  3  9  value  of  ditto  at  7s.  6d.  per  yard. 
Or  thus, 
|6|}|12     8     6  value  of  248^  yards,  at  Is.  per  yard. 
Multiply  by  7 

86  19     6  value  of  ditto  at  7s.  per  yard. 
6     4     3  value  of  ditto  at  6d.  per  yard. 

Ans.  i:93  3     9 


5  0 


By  the  latter  part  of  this  case, 
£      s.   d. 
248  10  0  value  of  2481  yard?,  at £1  per  yard. 


2  6  1    62     2  6  value  of  ditto  at  5s.  per  yard. 

31      1  3  value  of  ditto  at  2s.  Qd.  per  yard. 


Ans.  J£93     3  9  value  of  ditto  at  7s.  6d.  per  yard. 


Questions. 

Yds.  s.  d. 

2.  681  at     4  6  per  yd. 

3.  124^  —    5  8 

4.  146    —  14  9 — 

6,  2181— .  12  6 


Answers. 

£     s.    d. 

15     8     3 

35  2  8 
107  13  6 
136   11     3 


CASE  VI. 

^.    Whyin  the  quantity  is  any  number  less  than  1000,  and  the  price  not 
hnvre  than  12d.pcr  yardySrc  :  Find  the  value  of  the  whole  quarir 

X 


170  PRACTICE. 

tity  at  Id  per  yard,  which  may  be  done  by  dividing  it  by  12,  men- 
ially, setting  dowu  the  quotient  only  in  pounds,  or  shillings,  or  both. 
Then  multiply  this  sum  by  the  pence  in  the  price  of  1  yard,  and 
the  product  will  be  the  answer. 

Examples. 
1.  What  will  759|  yards  cost,  at  7d.  per  yard  ? 
s.     d. 
63     31  value  at  Id.  per  yard. 

Or,  £3'   3    3}  value  at  Id.  per  yard. 
Multiply  by  7 


Ans.  £22     3     0^  value  of  769i  yards,  at  7d.  per  yard. 
Questions.  Answers. 

Yds.  d.  £      5.      d. 

2.  975|  at  2    per  yard.     8     2     7 

3.  846    —  3-1    12     6     9 

CASE  VII. 

When  the  price  is  at  any  of  the  rates  in  the  second  Practice  Table 
of  aliquot  parts :  Multiply  the  given  quantity  by  the  numerator, 
and  divide  that  product  by  the  denominator  ;  if  the  price  be  pence, 
the  quotient  will  be  the  answer  in  shillings ;  if  shillings,  the  an- 
swer will  be  pounds. 

Examples. 
1.  What  will  379  yards,    at        2.  What  will  149  yards,  at  6s> 
4id.  per  yard  come  to  ?  per  yard  come  to  ? 

379  149 


3 

8)1137  1)0)44|7 


0 


2|0)14|2     1-t  Ans.  £44   14 

Ans.  £7  2   U- 

Questions.  Answers. 

Yds.  s.  d.  £    s.    d. 

3.  127    at  0  71  per  yard.  3  19     4| 

4.  159  —  0  8       6     6     0 

5.  173  —  0  9       6     9     9 

(].     241  —  0  10      10     0   10 

7.  249  —  7  6       93     7     6 

8.  357  —  12  6      223     2     6 

CASE  VIII. 

When  the  price  is  any  even  number  of  shillings,  if  it  be  required  to 
know  what  quantity  of  any  thing  rnay  be  bought  for  so  much  money  : 
Annex  a  cypher  to  the  money,  and  divide  it  by  half  the  price,  and 
the  quotient  will  be  the  quantity  to  be  purchased. 


PPAchCE. 


17 


Examples, 

1.  How  many  yards  of  cloth,  at  18s.  per  yard,  may  I  have  for 
345? 

Half  the  price  =^9)3450==:money  with  a  cypher  annexed. 


i;345 


3831  yards,  Ans  * 
Questions. 

2.  How  many  yds.  at  2  per  yd.  for  427  ? 

3.  4  312 

4.  6  917 
6.                                   8  195 


Answers. 

Yds. 
4270 
1560 
3050| 
487i 


CASE  IX. 

To  find  the  value  of  goods  sold  by  particular  Quantities,  viz.  I.  By 
the  score.  H.  Round  timber.  III.  By  5  score  to  the  hundred. 
IV.  By  112  to  the  hundred.  V.  By  6  score  to  the  hundred.  VI. 
By  the  great  gross^     VII.  By  the  thousand. 

I.  To  ^nd  the  value  of  goods  sold  by  the  score. 

The  price  of  one  is  given,  to  find  the  price  of  one  score. 

If  the  given  price  be  shillings  and  pence,  or  only  pence,  divide 
the  given  price,  in  pence,  by.  12.  The  quotient  will  be  the  an* 
swer  in  pounds,  and  the  remainder  will  be  so  many  times. Is.  8 J. 
for  a  score  is  20,  and  20d.  is  Is.  8d. 

Examples. 

1.  At  9d.  each:  What  is  that 
per  score  ? 


12)9d.(-75=£0     15    0  Ans. 

Or  by  inverting  the  question. 
1  score=20=ls.  8d. 
9 


2.  At  4s.  9d.  each:     What  is 
that  per  score  ? 

£4  15  Ans. 


15s. 0 


It  may  be  remarked,  that  when  the  price  is  shillings  and  pence, 
the  answer  will  be  just  so  many  pounds  as  there  are  shillings,  and 
so  many  times  Is.  8d.  as  there  are  pence.  If  farthings  are  given, 
for  ^d.  reckon  6d.  for  |d.  lOd.  and  for  Jd.  Is.  3d. 


2  is  tV 

5-i 


IQUOT    PARTS.       20    THE    INTEGER. 

12  is  f\ 

14 -t\ 
15-  -J 

16  is  A 
18  -  /^ 

M 


*  At  £1  a  yard,  the  quantity  would  be  345yds.  But  as  the  price  is  18s.  you 
want  only  l.8:=_9_^  of  the  qunntity.  This  explains  the  rule ;  for  the  course  is 
similar  /or  any  other  case. 


172  PRACTICE. 

3.  What   cost  7  ;    at  2s.    9d. 
per  score? 

s.    d. 


2     9  4.   What  cost  17;  at  19s.  lOd, 

per  score  ? 

0     8i  16s.  lOid.  Ans. 

0     3i 


7=0  111 

II.  Round  Timber. 

Forty  feet  make  a  load  or  ton  of  round  timber. 
If  the  given  price  of  a  foot  be  shillings, 

Rule. 

Multiply  the  given  price  by  2,  and  the  product  will  be  the  an- 
swer in  pounds. 

5.  What  cost  a  ton  at  3s.  per  6.  What  cost  a  ton  at  9s.  per 
foot?  3s.x2=61.  Ans.*        foot?  9s.x2=181.  Ans. 

If  the  given  price  of  1  foot  be  pence  only,  or  shillings  and  pence, 
divide  the  given  price,  in  pence,  by  6.  The  quotient  will  be  the 
answer  in  pounds,  and  the  remainder  will  be  so  many  times  3s.  4d. 

7th.  What  cost  40  feet,  at  17d. 
per  foot?  €th.  At  Is.  9d.  per  foot:  What 

6)17  cost  a  ton? 

£3  10  Ans. 

£2  16  8  Ans. 

If  the  given  price  of  a  foot  be  farthings  only,  or  pence  and  far- 
things, divide  the  given  price  in  farthings,  by  6  ;  then  divide  ihat 
quotient  by  4,  and  this  last  quotient  will  be  the  answer. 

9th.  At  '|d.  per  foot  :  What 
cost  a  ton  ? 

6)3  10th.  At  131  per  foot:    What 

cost  a  ton  ? 

4)0  10  £2  4  2  Ans. 

£0     2  6  Ans. 


♦  If  a  ton  of  timber  had  boon  20  feet,  then  3s.  a  foot  would  have  mauo  £.Z ; 
but  as  a  ton  has  twice  20  feet,  the  answer  in  pounds  must  be  twice  the  number 
of  shiUings.     In  like  manner  for  any  number  of  shillings.     If  the  price  be  pence 

40X17     20X17     17 

a  foot,  as  in  Ex.  7,  the  rule  is — =: = — for  pound?,  witli  a  remainder 

12  tj  6         ^ 

209. 
of  5X  -7-  =5  times  2s.  4d. 
tl 

40x3 
When  the  price  is  farthings  a  foot,  the  rule  is  as  in  Ex.  9, for  shillicg.?, 

20-f-3       3     .  , 

^^ for  pound?. 

0X4      6X4        *^ 


PRACTICE.  173 

Or,  suppose  every  shilling  in  the  price  to  be  21.  every  penny  to 
be  3s.  4d.  and  every  farthing  to  be  lOd. 

12lh.  What  cost  40  at   16id. 
per  foot  ? 
nth.  What   cost  40   feet   at  s»  d. 

|d.  per  foot?  lOx    2=£2     0  0 

|d.  X  10  £0  2  6  Ans.  3  4  x    3=    0  10  0 

0  4-  X  10=    0     1  8 


£2  n 


III.*  To  find  the  value  of  goods  sold  hy  5  score  to  the  hundred. 
1st,  If  the  given  price  be  pounds  and  shillings,  or  shillings  only. 

Rule. 
Multiply  the  given  price  in  shillings,  by  5,  and  the  quotient  will 
be  the  answer  in  pounds,  for  lOOs.  are  £5. 

13th.  At   19s.  per  yard,  what 
cost  100  yards  ?  14th.  At  41.  135.  per  cwt.  what 

193,  cost  100  cwt.  or  5  tons  ? 

5  '      J£465  Ans. 

£95  Ans. 
2d.  If  the  given  price  of  1  be  pence  only,  or  shillings  and  pence. 

Rule. 
Multiply  the  given  price  in  pence,  by  5  ;  then  divide  that  pro- 
duct by  12.     The  quotient  will  be  pounds  ;  and  the  remainder  so 
many  times  Is.  8d. 

16th.     What  cost  100  bushels, 
at  35s.  4d.  per  bushel  ? 
15th.  Ifl   yard  cost  9d.  what  s    d.  Or, 

cost  100  vards  ?  35  4  35s.  4d. 

'9  J^  J4d.|i|  5 

—  424  175 

12)45  5  1    13  4 


£3  15  Ans.  12)2120  £176   13  4 


£17G   13  4  Ans.     Here  5  is  di- 
vided  by  i. 


*  In  Federal  Money. — Remove  the  decimal  point  two  places  to  the  right  for 
the  answer. 

Examples. 

1.  What  cost  100  yards  at  $2  50c.  per  yard.?  $2-50 X  100=$250-  An?. 

2.  What  cost  100  yards  at  75c.  per  yard  ?  $-75  X  100=:$75-  Ans. 
•3.  What  cost  100  yards  at  5c.  64m.  per  yard .? 

$•05625  X  100=$5-625  An*. 
4.  What  cost  100  yards  at  37c.  5m.  per  yard  ?  Ans.  $37  50c. 

6    What  cost  100  yard=*  at  68c.  7ira.  per  yard  ^  Ans.  $68  75c. 


174  PRACTICE. 

3.  If  the  given  price  of  1  be  shillings  and  pence  ;  Multiply  the 
price  by  6,  and  the  product  under  the  place  of  shillings,  will  be  the 
answer  in  pounds,  and  the  product  under  the  place  of  pence,  will 
be  so  many  times  Is.  8d. 

17th.  At  2s.  6d.   per  bushel: 
what  cost  100  bushels  ? 
s.  d. 

2  5  18th.    At   25s     3d.   per    ton  ^ 

5  what  cost  100  tons  ? 

£126  5  Ans, 

12  1 


£12   1  8  Ans. 
4.*  To  find  the  price  of  owe  at  so  much  per  hundred  of  5  score. 

General  Kule. 
Multiply  the  given  price  by  12 ;  divide  the  product  hy  5,  ant! 
the  quotient  will  be  the  answer  in  pence. 

But  if  the  price  be  pounds  only  ;= 
Kule. 
Divide  the  given  price  by  5,  and  the  quotient  will  be  the  answer 
in  shillings. 

19th.  If  100yds.  cost  £6ii,  what        21st.  If  100  yards  cost  £11 '^. 
cost  1yd?  9d.  what  cost  1  yard. 

5)65  £     s.     d, 

—  11     7     9 

13s.  Ans.  12 


20th,  If  lOOyds.  cost  £2   18s.  5)136   13     a 

4d.  what  is  that  per  yard  ?  

£   s.    d.  12)27     6     7 

2   18  4  ' 

12  2s.  3id.  Ans. 

In  dividing  27  by  12  (in  the 

5)35  0  0  21st  question)  the  quotient  is  2s- 

and  the  remainder  3d.  the  6  is. 

7d.  Answer.  2^^-  of  a  penny  =  one  farthing, 

■ —  and  the  7  is  of  no  account. 

TARLE    OF    ALIQ;UOT    PARTS.        100  THE    INTEGEPw 

5     is     A!25     is 


10 ,VPO 3 

20 


50     is      I 

60 A 


75     is      I- 

80 -,\ 

90 /o 


*'  In  Federal  Money. — Renicre  (.lie  decimal  poiiit  two  places  to  tlie  left  for 
the  answer. 

Examples. 
I.  Tf  100  yards  cost  |i>50,  v.rhat  cost  1  yard .?  $250~100=::$2v>0  Ans. 

-2.  in 00  yards  cost  $7b'  what  cost  1  yard  .=>  $75'~100==$-75  Ans. 

3.  If  100  yards  cost  $3  62c.  5m.  what  cost  1  yard  ? 

$5-625-M00~|-05C25=r5c.  6i^.  Ans. 

4.  M  100  yards  cost  .$37  50c.  what  cost  1  yard.^  Ans.  37c.  5m. 

5.  If  100  yards  cost  |ca  75e.  what  cost  1  yard  ?  Ans.  68o.  7^m. 


PRACTICE. 


175 


22d.*  At  £3  7s.  6d.  per  100  :  What  will  23  cost  ? 


20 


3     7  6 


0  13  6 

0     1  4  >  Add 

0     0  8 


23=  £0  15  6  Ans. 
23d.  At  £2  Is.  lOd.  per  100  :         24lh.  At  £5  9s.  6d.  ner  100 i 


What  cost  18? 

je  s.  d. 

20 


* 

2   1    10 

I  0 

0  8     4f 
0  0  10 

What  cost  35  ? 

£     p.     d. 

5     9     6 

3 


Sub. 


18=  £0  7     6fADs. 


10)16     8     6 
6 


1    12   10 
5     6f 


Add. 


£1   18     3^  Ans. 


IV.   To  find  the  value  of  goods,  sold  by  1 12ife  the  Cwt. 

The  price  of  life  is  given  to  find  the  value  of  1  cwt. 

Rule. 
For  a  farthing,  account  2s.  4d.  per  cwt.     For  a  half  a  penny, 
4s.  8d.     For  three  farthings,  7s.     And  for  every  penny  9s.  4d. 
per  cwt. 

25th..  What  cost  Icwt.  at  3id.         26th.  At  8|d.  per  ft :   What 
per  ft  ?  cost  1  cwt  ?      Ans.  £4  Is.  8d, 

At  Id.  per  ft  s.  d. 

Icwt.  costs  9  4 


At  3d.       £18  0 

At  id. 


1     8  0^ 
0     4  85 


Add. 


£1   12  8  Ans. 


*  To  find  the  value  of  any  number  at  a  given  price  per  100,  in  federal  money,— 
Multiply  the  price  per  100  by  the  given  quantity,  and  point  oft"  two  right  hand 
figures,  in  the  product  more  than  required  by  multiplication  of  decimals.  Or, 
point  off"  the  two  right  hand  places  in  the  given  quantity,  and  multiply,  and 
point,  as  in  multiplication  of  decimals. 

EXAM-PLES. 

t.  What  cost  56  yards  at  $87  50c.  per  100  yards  ? 

$87-5  xsa 

^~— 149,  Ans,    Or,  $87*5  X  "56=i549,  as  before. 

iOO 


ivtJ 


PRACTICE. 


V.   To  find  the  value  of  goods  sold  by  6  score,  to  ike  hundred. 
The  price  of  1  is  given  to  find  the  price  of  1  hundred. 

Rule. 
Suppose  every  penny  in  the  price  to  be  so  many  pounds,  and  for 
the  farthings,  such  a  part  of  a  pound,  as  they  are  of  a  penny  j 
then,  half  of  that  sum  will  be  the  answer. 

27th.  At  4-td.  per  yard  :  What         28th.  At  16s.  Ojd.  per  yard  : 
cost  120  yards  ?         "  What  cost  120  yards. 

£  s.  Ans.jeiao  12s.  6d. 

2)4   10 

£2    6  Ans. 
To  find  the  price  of  one,  at  so  much  per  hundred  of  6  score. 

Rule. 
Blulliply  the  price  by  2,  then  call  the  pounds  so  many  pencCj 
and  the  shillings,  such  a  part  of  a  penny,  as  they  are  of  a  pound, 
and  you  will  have  the  answer. 

29th.  If  120  yards  cost i:3  12s.:         30th.  If  120  yards  costce^/ 
What  cost  1  yard  ?  18s.  6d.  :   What  cost  1  yard'' 

£  8.  Ans.  ll^d.+l  of  farthing. 

3  12 


7     4 


Ans.  7id. 


TABLE    OF    ALIQUOT    PARTS.        120    THE    INTEGER. 


Also, 


6  is   2V 

10  -  -,v 

12  -  tV 

20—   A 


24  is  1 

36  is  j\ 

72  is    1 

30  is  1 

45  -  1 

75—  f 

40  is  i 

48—  1 

80-1 

60  is  1 

50  -  j% 

84 -^V 

70-^^ 

90—  f 

96 
100 
105 
108 


,4 

5 


IS    4 


:Ust.  At  £3  17s.  6d.  per  hundred,  what  cost  14 
£  s.   d. 
12  j\  3  17   e 


0     1  31  P^^^- 


14=  £0    9  Oi  Ads. 


i>.  What  cost  45 ^Ib.  beef  at  $5i  per  100-' 
4f^.y  v'45'5 

-=$2-5025=$2  50c.  2hm.  Ans.  Or,  $5-5 X  •455lb.=$S'5025,  as  before- 

100 

3.  What  cost  375  yards  at  $375  per  100  yards?  Ans.  $1404  25c. 

4.  What  cost  54  yards  at  $16  per  100  ?  Ans.  $8  64p. 
■}.  What  cost  512  vards  at  $6  25c.  per  100  yards  ?  Abb.  p^ 


J 


PRACTICE. 


Ill 


32.  At£2  135.  6id.  per  hua-        33.  At£l   19s.  3(1.  per  hun- 
dred, what  cost  49  ?  dr^d,  what  cost  75  ? 

£   s.     d.  Ans.JSl  4s.  6Ad. 

40 


2   13     61 


0  17  10  0?- 
0  3  6  Si 
0     0     5   1" 


49  =£1     1   10  1  Ans. 

VI.  !ro  y?rtrf  <Ac  value  of  goods  sold  by  the  great  gross. 

Note.  12  make  1  dozen,  12  dozen  1  small  gross,  12  small  gross 
1  great  gross. 

The  price  of  1  dozen  being  given,  in  pence,  to  find  the  price  of 
a  great  gross. 

Rule. 

Multiply  the  price  of  1  dozen,  in  pence,  by  3,  then  divide  that 
product  by  5,  and  the  quotient  will  be  the  answer  in  pounds,  &c. 
For  proof,  do  the  contrary. 

N.  B.  If  the  price  of  1  be  given,  the  price  of  1  small  gross  is 
found  after  the  same  manner. 

34.  What  cost  1  great  gross,  at  18d.  per  dozen? 

3 

5)54 

£10     16 

35.  At  4s.  3d.  per  dozen,  what  cost  1  great  gross  ? 


4s.  3d. 
12 

61d. 
3 


Or, 

s.  d. 

4  3 

12 


Or, 
£s. 

144=7  4 
4 


5)153 


2   11   0 
12 


\''-\i\'l     |«jAdO, 


Ans.  £30  12 


£30  12  0 


£30     12 


TABLE    OF    ALIQUOT    PARTS.        144    THE    INTEGER. 


Also, 


12  is 

16  — 


1 

i 

18 -i 
24-1 


36  is  1 

32  is    f 

84  is  r\ 

48  -  1 

60  ~  ^V 

96—1 

72  — i 

34-1- 

108  —  1 

80—  1 

120  ^  -^ 

32 


178 


PRACTICJL. 


36.  At  £2  12s.  9d.  per  great  gross,  what  Cost  45  dozen. 
Doz.      £  s.    d. 


i 

2  12  9 

i 

0  13  21 
0  3  31 

Add. 


46  =£0  16  6f  Ans. 

37.  What  cost  117  dozen,  at        38.  At  £3  16s.  8d.  per  great 
£9  13s.  7d.  per  great  gross?         gros^,  what  cost  7    great  gross' 


Doz. 

!08 


£    s.  d. 

And  96  dozen  ? 

9  13  7 

£   s.   d. 

3 

3  16  8 

1 

29  0  9 

tV 

7  6  2 
12  1 

n 


3)7   13  4 


Add. 


117  =  £7  17  3J  Ans. 


Doz. 
96=  2   11    1{ 
top  line  X7=26   16  8 


£29 


7  91 


VII.*  To  find  the  value  of  goods  sold  by  the  thousand. 
The  price  of  1  is  given  to  find  the  price,  of  1000. 

RuLE-t 

Multiply  the  given  price  in  pence,  by  60,  then  divide  the  pro- 
duct by  12,  and  the  quotient  will  be  the  answer  in  pounds,  &c. 

39.  At  6d.  each  ;      Or,  as  1000s.  are  £60,  40.  What  cost 

what  cost  1000  ?  take    parts,   for  the  1000,    at    2id. 

6  pence  out  of  50.  each  ? 

60  |6d.|i|50  Ans.£9  7s.  6d, 


12)300 
£25  Ans. 


Ans.  25 


*  In  Federal  Money. — Remrive  the  decimal  point  three  places  to  the  right,  or 
left,  as  the  case  requires,  for  the  answer. 

Examples. 

1.  What  cost  1000  yards  at  5  cents  per  yard?     -05  X 1000— 050-=:.$50  An?. 

2.  What  cost  1000  yards  at  12  cents  5  mills  per  yard  ?  Ans.  $125. 

3.  If  1000  yards  cost  $37  50c.  what  cost  1  yard. 

$37-5-MOO0— $-0375=3c.  7^m.  or,  3|c.  Ans. 

4.  If  1000  yards  cost  $1625,  what  cost  1  yard  ?  Ans.  $1  62c.  Sm. 

t  The  operation  by  the  Rule  of  Three  would  be  thus ;  in  the  first  example, 

6  X  1000 
as,  1  :  6d.  ::  1000  :  GxlOOO  the  answer  in  pence;  and —rr — j^-r— answer  in 

6X50 
pounds,  =  — T-p  pounds,  and  is  the  rule. 


PRACtlCE. 


179 


VIII.  To  And  the  price  of  on^  at  so  much  per  thousand. 

HULE. 

Multiply  the  price  by  12  ;  divide  the  product  by  60  ;  then  take 
*he  pounds  for  so  many  pence,  and  the  shillings  for  s«ch  a  part  of 
a  penny  as  they  are  of  a  pound,  which  will  be  the  answer. 
41.  AX  £5  4s.  2d.  per  1000,  what  cost  J? 
£    s.  d. 
5     4  2 
12 


50 


6)62   10  0 
10)12   10 


£1 


42. 
what 

At  £3 
cost  1 

50  j 

Ans. 
54  3s.  4d.  per 

? 

£      8,  d. 

364  3  4 
12 

IQOQ, 

10)4250  0  0 

5)425 
85 
An|.7s.  Id. 

00 

tV 

Or. 
£ 

3^4 

s. 
3 

d. 
4 

10 

35 

8 

4 

1 

3 

10 

10 

Ans.  0     7     1 


TABLE    OF  ALIQUOT  PARTS.        1000  THE.   INTEGER. 

700   is    j\ 


60  is  2^1 

200  is  i 

100  —  j\ 

250  —  1 

125  —  i 

300  --  J- 

300 

is 

A 

375 

— 

# 

400 

— 

f 

600 

— 

1 

625 

— 

f 

800  —  |- 
875-  J 


900  --  A 


*At£l  17s.  9d.  per  1000  what  coat  115? 


Ads^. 


£ 

s.  d. 

100 

1 

17  9 

10 

0 

3  9i 

5 

^ 

0 

0  4i 

0 

0  2i 

115=£0    4  4  Ans, 


*  To  find  the  value  of  any  number^  at  a  given  price  per  1000  in  federal  mo-' 
ney. — Multiply  the  price  per  1000  by  the  given  quantity,  and  poiiit  off  three 
i'ight  hai^d  figures  in  the  product  more  than  required  by  multiplication  of  deci- 


180 


PRACTICE. 


44th.  At  £2  Is.  8d.  per  1000,        45th.    What  cofet  33,  at  24s. 


what  cost  875  ? 

£     8.     d. 

2     1     8 
7 


8d.  per  1000  ? 


8)14  11     8 


£1   16     S^Abs. 


50 

2  0 

25 

i 

5 

i 

30 



3 

t'o 

33 

= 

£s.  d. 
1   4  8 


0   1   2| 


0  0  7,^ 
0 


0  0  8i>  .  ,, 

0  0  |r^^' 


0  0  9f 


CASE  X. 

When  the  price  of  1  is  any  number  of  dollars  and  parts  of  a  dollar : 
Multiply  the  quantity  by  the  number  of  dollars  ;  and,  finding,  by 
the  general  rule,  the  price  at  the  parts  of  gl,  the  sum  of  the  \YhoIe 


is  the  answer 

Examples. 
1.  What  cost  395  yards  at  g3  24c.  per  yard  ? 

c.  k    c. 


20 


of  20c. 


395 
3 

1185 
79 
15 


=  price  al.  gl 


=  ditto  at 

=  ditto  at 

80  =  ditto  at 


20c. 
4 


Ans.  $1279  80  =  ditto  at    ^3  24c. 

2.  What  cost  269  yards  at  $2  60c.  per  yard  ?  Ans.  ^699  40c, 

3.      694    12   10  8397  40 

4.      318    4   121 1311   75 

5.      175    4  44 .  777 


CASE  XL 

When  the  price  of  1  contains  the  same  aliquot  part  of  a  dollar  antj 
number  of  times  exactly ;  or,  in  other  -wordsy  when  the  price  of  1  has 
an  aliquot  part,  which  is  also  an  aliquot  part  of  a  dollar  :  First,  {iml 
the  value  of  the  given  quantity  at  the  aliquot  part  j  then  multiply 

KiJils.     Or,  point  off  the  three  right  hand  places  in  the  givejo.  quantity :  and 
multiply  and  point  as  in  multiplication  of  decimals. 

Examples. 

1.  Vv'hat  cost  875  at  $13  per  lOCO  ? 

875X13=11375;  and  n375-?-1000=n-375=$n  37c.  5m.  Ans. 
'2.  What  cost  39175  feet  of  boards,  at  $16  per  1000  ?  Ans.  $626  OOo. 

•;.  What  cost  325  nails  at  $1  50c.  per  1000  .^  Aus.  48c.  "l^m.,  or,  48{q, 


PRACTICE.  18! 

this  by  the  number  of  times  which  the  aliquot  part  is  contained  in 
the  given  sum,  for  the  answer.     Or, 

Since  the  price  in  this  case  is  alwa;ys  such  a  number,  as,  being 
divided  by  the  aliquot  part,  will  make  the  numerator  of  a  fraction, 
of  which  the  denominator  is  the  denominator  of  that  fraction,  which 
the  aliquot  part  is  of  a  dollar  ;  Multiply  the  quantity  by  the  nu- 
merator, and  divide  the  product  by  the  denominator,  (or,  when 
convenient,  divide  the  quantity  by  the  denominator,  and  multiply 
the  quotient  by  the  numerator,)  for  the  answer.* 


Examples. 
J. 

2 


1.  What  cost  384  yards  at  87|  cents  per  yard  ? 
12ic.  =  ^  of  -875  =  I  $i  I  384-  =  price  at     $1 

48-  =F  ditto  at  -12^ 

X7  X  7*^ 

Ans.  $336-  =  ditto  at  -37^ 

Or  thus, 

.875=31,  and  384x|=='«V'^'(=='f*X7)— g336,  Ans.  as  before 

2.  What  cost  842  yards  at  66|c.  per  yard  ?     Ans.  §561   33^c. 

3.  What  cost  912  yards  at  55c.  per  yard  ?        Ans.  g501  60c. 

MISCELLANEOUS    QUESTIONS    IN    PRACTICE. 

I.  What  cost  300  yards  at  27c.  per  yard  ?  Ans.  g81 

917           $1   12  5m.  1031  62c.  5in. 

351       35  12  32 

862^  ft.  boards  at  gl2  per  M.  ?  10  34  6 

32159            13  75c.  442  18  6^ 

CASE  XII. 

L  Since  2s.  is  ^V  o^  *^^'  the  decimal  of  2s.  is  •!  :  Wherefore  any 
quantity  being  given  at  2s.  per  lb.  yard,  &c.  the  price  is  found  in 
pounds  and  decimal  parts  of  a  pound,  by  separating  the  unit  figure 
of  the  given  quantity  from  the  rest,  for  a  decimal. 

Let  it  be  required  to  find  the  value  of  356  yards  at  2s.  per  yard  ^ 
By  pointing  off  the  unit  figure  6  for  a  decimal,  I  find  the  >    £oc.f; 
_.  lount  to  be  £356,  which  is  known  to  be  equal  to  351.  12s.  ^ 

II.  Consequently,  if  the  price  be  a  rnultiple  of  2s.  (viz.  any  even 
number  of  shillings)  the  amount  at  2s.  being  first  found  in  pounds 
and  decimal  parts,  as  above,  and  that  amount  multiplied  by  the 
number  which  shows  how  often  2s.  is  contained  in  the  given  price, 
the  product  will  be  the  amount  required  in  pounds  and  deci-mal 
parts  of  a  pound. 

What  cost  427  gallons  of  wine,  at  8s.  per  gallon  ? 

*  Some  of  the  prices  wliich  apply  to  this  case,  are  to  be  found  in  the  seconU 
table  of  parts  of  a  dollar. 


amo 


182 


PRACTICE. 


£42  7  amount  at  2s.  per  gallon. 
4 


Ans.  £170-8  or  1701.  16s. 
The  examples  in  Case  1st.  may  be  worked  in  this  manner. 
Likewise,  if  the  price  be  pounds  and  even  shillings. 
754  yards  at  11   8s. 


76-4  amount  at  2s. 
14x2=28s. 


3016 
764 


Or, 

734 
75-4x4=301-6 


Add. 


£l05d-6 


Ang.  £1065-6=]0551.   12s. 

III.  If  the  price  be  an  aliquot  part  of  2s. :  Find  the  amount  at  2s. 
»nd  divide  it  by  the  denominator  of  the  part,  and  the  quotient  wiU 
be  the  answer. 

At  8d.  per  lb. :  What  cost  976  lb.? 


I  8d.  U  I  97-6 

£32  533=  £32     10     8  Ans. 
IV.  If  the  price  be  not  an  aliquot  part :    Divide  it  into  aliquot 
parts. 

7235  yards,  at  7d. 


4d. 
3d. 


723'5 

120  583 
90-437 


211-02  =£211     0     4|  Ans. 

V.  If  the  price  be  pounds  and  shillings,  or  pounds,  shillings  and, 
pence:    Reduce  the  shillings,  &c.  to  the  decimal  of  a  pound,  and 
multiply  the  quantity  thereby,  or  the  price  by  the  quantity. 
At  151.  12s.  6d.  per  Cwt :  VVhat  cost  75  Cwt.  ? 
£15  12  6  =  £1^-625 
75 


78125 
109375 

1171-875 


£1171   17  6  Ans. 

'  VL  Jf  the  quantity  likewise  be  of  divers  denominations  :    Reduce 
the  less  denominations  to  the  decimal  of  that,  whereof  the  orice  is 


PRACTICE. 


n$ 


:^b.  lOoz.  of  silk,  at  £4  5  9  =£4-287 
mb.  lOoz.  =9-625 


21435 
8574 

25722 
38583 

41-262375 


£415    3  Ans, 

teases  6th.  and  7th.  may  be  wrought  io  this  manner. 
Or,  you  may  take  parts  for  the  lower  denominations. 


8oz. 
2oz. 


4-287 
9 


38-583 
2-1435 
•535875 

41-262375 


£41 


VII.  When  the  price  is  any  odd  number  of  shillings  :  h  it  be  re- 
tq^uired  to  know  what  quantity  of  any  thing  may  be  bought  for  any 
sum  of  money,  in  pounds  :  Annex  two  cyphers  to  the  money,  and 
divide  it  by  half  the  price. 

Note.  As  half  a  shilling  (or  6  pence)  is  '5,  therefore,  to  halve 
any  odd  number  of  shillings,  is  only  to  annex  '5  to  half  of  the  great- 
est even  number  in  the  price. 


1st.  How  many  yds.  at  7s.  per 
yd.  may  I  have  for  4351.  ? 

Half=3-5)43500(1242||-yds.Ans. 
35 

85 
70 


150 
140 


2d.  How  many  pounds  of  tea, 
at  5s.  per  lb.  for  £37? 

1481b.  Ans. 
3d.  How   many   yards   at   9s. 
per  yard  may  I  have  for  5401.  ? 
'An^,  1200  yards. 


100 
70 


30 


14  TARE  AND  TRET. 

BlLL  OF  PARCELS. 

Newburyport,  January  1st,  1808. 
Mr.  Timothy  Huckster 

Bought  o{ Samuel  Merchant, 

?5|^ife  Bohea  tea,  at  3s.  6(1.  per  ft. 
48ft  Cheese,  at  9d.  per  ft. 
15  Pair  worsted  hose,  at  6s.  8d.  per  pain 
4^  Dozen  vvomen's  gloves,  at  36s.  6d.  per  dozen. 
19  Dozen  knives  and  forks,  at  6s.  9d.  per  dozen. 
9  Grindstones  at  15§.  9d.  per  stone. 
^  Cwt.  Brown  sugar,  at  51s.  percwt. 

31  ft  Loaf  Sugar,  at  Is*  O^d.  per  ft.  — . 

£34     3 

.   Received  payment  in  full,  

Samuel  Merchant, 


TARE  AND  TRET. 

TARE  and  Tret  are  practical  rules  for  deducting  certain  allow- 
ances,  which  are  made  by  merchants  and  tradesmen  in  selling  their 
goods  by  weight. 

Tare  is  an  allowance,  made  to  the  buyer,  for  the  weight  of  the 
box,  barrel  or  bag,  &c.  which  contains  the  goods  bought,  and  is 
either  at  so  much  per  box,  &c»  at  so  much  per  cwt.  or  at  so  much 
in  the  gross  weight. 

Tret  is  an  allowance  of  4ft  in  every  104ft  for  waste,  dust,  kc. 

Cloff  is  an  allowance  of  2ft  upon  every  3cwt. 

Gross  Weight  is  the  whole  weight  of  any  sort  of  goods,  together 
with  the  box,  barrel,  or  bag,  &c.  which  contains  them. 

Sutlle  is,  when  part  of  the  allowance  is  deducted  from  the  gross. 

Neat  weight  is  what  remains  after  all  allowances  are  made. 

CASE  I.* 

When  the  tare  is  at  so  much  per  box.^  barrel  or  bagy  ^'C.  :  Multiply 
the  number  of  boxes,  barrels,  &c.  by  the  tare,  and  subtract  the 
product  from  the  gross,  and  the  remainder  will  be  the  neat  weight 
required. 

ExAMlPLES. 

1.  In  6  hogsheads  of  sugar,  each  weighing  9cwt.  2qrs.  lOlb.  gross, 
tare  251b.  per  hogshead  ;  how  much  neat  ? 

*  This,  as  well  as  every  other  case  in  this  rule,  is  bnly  an  applieation  of  tb^ 
rales  of  Proportion  and  Practice. 


TARE  AND  TRET.  185 

Cwt.  qr.    ft         Cwt.  qr.  ft 
25x&=l     1     10  9     2  10  gross  wt.  of  Ihhd. 

6 


57     2     4  gross. 
1     1   10  tare. 


Ans.  56     0  22  aeat. 
2.  In  5  bags  of  cotton,  marked  with  the  gross  weight  as  follows, 
tare  23ft  per  bag  ;  what  neat  weight? 


Cwt. 

qr.  ft 

A=7 

1   19 

B=3 

3  2l 

C=5 

1   12 

D=6 

0  13 

E=8 

1     0 

Cwt.  qr.  ft 

— 

Ans.  30  0  14  neat. 

3.  What  is  the  neat  weight  of  15  hogsheads  of  tobacco,  each 
7cvvt.  Iqr.  13ft,  tare  100ft  per  hogshead? 

Ans.  97cwt.  Oqr.  lift. 

CASE  II. 

When  the  tare  is  at  so  much  per  cwt.  :  Divide  the  gross  weight  by 
the  aliquot  parts  of  a  cwt.  subtract  the  quotient  from  the  gross, 
and  the  remainder  will  be  the  neat  weight. 

Examples. 

1.  In  129cvvt.  3qrs.  161b.  gross,  tare  14Ib.  per  ewt.  what  neat 
weight  ? 

Cwt.  qr.   lb. 
141b.  I  i  I  129     3     16    gross. 
I     16     O     26i  tare. 


Ans.   113     2     171^  neat. 

2.  In  97cwt.  Iqr.  71b.  gross,        3.  What  is  the  neat  weight  of 

tare   201b.  per  cwt.   what    neat    9  barrels  of  potash,  each  weigh- 

weight.  ing  3051b,  gross,  tare  121b.  per 

Ans.  79Cwt.  3qrs.  211b.  neat.    cwt.  ?      Ans.  2450ft  14oz.  44dr. 

4.  What  is  the  value  of  the  neat  weight  of  7hhd8.  of  tobacco,  at 

51.  7s.  6d.  per  cwt.  each  weighing  8cwt.  3qrs.  10ft  gross,  tare  21ft 

per  cwt.  ?  Ans.  £270  4'  4}  reckoning  the  odd  ounces. 

CASE  HI. 

When  tret  is  allowed  with  tare  :  Divide  the  suttle  weight  by  26, 
and  the  quotient  will  be  the  tret,  which  subtract  from  the  suttle, 
and  the  remainder  will  be  the  neat.-^ 

*  Tret  is  41b.  in  104,  which  is  -_4_=  i  .  And  CIofF  i?  51b.  in  3€wt.  or 
•  Hh.  which  IS  _|-=_J_. 

Z 


\m  INVOLUTION. 

Examples. 
1.  In  247cu;t.  2qt:s.   15il5  gross,  tare  281fe  per  cwt.  aiid  tret  4fe 
for  every  104!fe  what  neat  weight  ? 

I  28  I  I  I  247C.2qr.l5ife  gross. 

61     3     17     12  tare,  subtract. 


I  4  I  2V  I  185     2     25       4  suttle. 


7     0     16       0  tret,  subtract. 


Ans.  178  2  9  4  neat. 
2.  What  Is  the  neat  vveight  of  4hh(ls  of  tobacco,  weighing  as  fal- 
low: The  1st.  5cvvt.  Iqr.  121fe  gioss,  tare  65ife  per  hhd. ;  the  2d. 
3cvvt  Oqr.  191fe  gross,  tare  75ife  :  the  3d  6cwt.  3qrs.  gross,  tare 
49ife  ;  and  the  4lh,  4cwt.  2qrs.  9fe  gross,  tare  35ife,  and  allowing  tret 
to  each  as  usual  ?  Ans.  17cwt.  Oqr.  19ife4- 

CASE  IV. 

When  tare,  tret,  and  cUrff^  are  allowed :  Deduct  the  tare  and  tret  as 
before,  and  divide  the  sattle  by  168,  and  the  quotient  will  be  the 
cloff,  which  subtract  from  the  suttle,  and  the  remainder  will  be  the 
neat. 

Examples. 
1.  What  is  the  neat  weight  of  Ihhd.  of  tobacco,  weighing  16cwt. 
2qrs.  20lb   gross,  tare  141fe  per  cwt.  tret  4ife  per  104,  and  cloflf2it 
per  3cwt.  ? 

14feisi)16     2  20     0  gross. 

2     0     9     8  tare,  subtract. 


43feis  ^\)U     2   10     8 

0     2     6   13  tret,  subtract. 


2ife  is  lis)^-*     ^     3^1  *"^^'^- 

0     0     9     5  cloff,  subtract. 

Ans.  13     3  22     6  neat. 
2.  If  9hhds.  of  tobacco/ contain  85cwt.  Oqr.  2ife,  tare  30ife  per 
hhd.  tret  and  cloff  as  usual,  what  will   the  neat  weight  come  to  at 
€^d.  per fe  after  deducting  for  duties  and  other  charges,  511.  lis. 8d  ? 

Ans.Jt:iiJ7   18s.  5d. 


INVOLUTION 

TEACHES  the  method  ofiinding  ihe  powers  of  numbers. 

A  power  is  the  product  arising  from  multiplying  any  number  in- 
to itself  contmnally  a  certain  number  of  times.  Thus,  any  number* 
is  called  the  ^rs^  power,  as  3  ;  if  it  be  multiplied  by  itself,  the  pro- 
duct is  called  the  second  power  or  square,  as  3x3  ;  if  the  second 
power  be  multiplied  by  the  first  power,  the  product  is  called  the 


INVOLUTION.  18-7 

i^rc/ power,  or  cube,  as  3x3x3;  if  the  third  power  be  multiplied 
by  the  first  power,  the  product  is  the  fourtk  power,  or  biquadrate, 
as  3x3x3x3,  or  81  is  the  fourth  power  of  3,  and  so  on. 

The  power  is  often  denoted  by  a  figure  placed  at  the  right  and 
a  little  above  the  number,  which  figure  is  called  the  index  or  ex- 
ponent of  that  power.  The  index  or  exponent  is  always  one  more 
than  the  number  of  multiplications  to  produce  the  power,  or  is 
equal  to  the  number  of  times  the  given  number  is  used  as  a  factor 
in  producing  the  power.     Thus  the  square  of 

3,  =  3  X  3  =  32  ;  and  the  cube  of 

3,  =3x3x3  =  33;  and  the  4th  power  of 

3,  =3x3x3x3  =  3*;  and  the  5th  power  of 

3,  =  3  X  3  X  3  x  3  X  3=3^ ,  and  so  on. 

In  producing  the  square  of  3,  for  instance,  there  is  only  one  mul- 
tiplication, or  two  factors  ;  ia  producing  the  cube,  there  are  two 
multiplications  or  three  factors,  and  so  on. 

Hence,  Involution  is  performed  by  the  following 

Rule. 

Multiply  the  given  number,  or  fi,rst  power  continually  by  itself, 
till  the  number  of  multiplications  be  1  less  than  the  index  of  the 
power  to  be  found,  and  the  last  product  will  be  the  power  required. 

Note.  Whence,  because  fractions  are  multiplied  by  taking  the 
products  of  their  numerators,  and  of  their  denominators,  they  will 
be  involved  by  raising  each  of  their  terhis  (o  the  power  required, 
and  if  a  mixed  number  be  proposed,  either  reduce  it  to  an  impro- 
per fraction,  or  reduce  the  vulgar  fraction  to  a  decimal,  and  pro- 
ceed by  the  rule. 

Examples. 

1.  What  is  the  5th  power  of  9  ? 

9 
9 

8I=2d.  power. 


729=3d.  power. 
9 

G561-— 4th.  power. 
9 

59049=5th  power,  or  ansvver=95. 

2.  What  is  the  5th  power  of  f  ?  Ans.  3^^"^,. 

3.  What  is  the  fourth  power  of -045  ?      Ans.  -000004100625. 
Here  we  see,  that  in  rai>iing  a  fraction  to  a  higher  power,  we 

decrease  its  value. 


188  EVOLUTION. 


EVOLUTION, 

OR  THE  EXTRACTION  OF  ROOTS. 

THE  Root  is  a  number  whose  continual  multiplication  into  itself 
produces  the  power,  and  is  denominated  the  square,  cube,  biqiia- 
drate,  or  2d.  3d.  4th.  root,  &c.  accordingly  as  it  is,  when  raised  to 
the  2d.  3d.  &c.  power,  equal  to  that  power.  Thus,  4  is  the  square 
root  of  16,  because  4x4=16,  and  3  is  the  cube  root  of  27,  because 
3x3x3=^:27,  and  so  on. 

Although  there  is  no  number  of  which  we  cannot  find  any  power 
exactly,  yet  there  are  many  numbers,  of  which  precise  roots  can 
never  be  determined.  But,  by  the  help  of  decimals,  we  can  ap- 
proximate towards  the  i^ot  to  any  assigned  degree  of  exactness. 

The  roots,  which  approximate,  are  called  surd  roots,  and  those 
which  are  perfectly  accurate,  are  called  rational  roots. 

Roots  are  sometimes  denoted  by  writing  the  character  v'  before 
the  power,  with  the  index  of  the  power  over  it ;  thus  the  3d  root 

3 

of  36  is  expressed  >/  36,  and  the  2d  root  of  36  is  V  36,  the  in- 
dex 2  being  omitted  when  the  square  root  is  designed. 

If  the  power  be  expressed  by  several  numbers,  with  the  sign  -f- 
or  —  between  them,  a  line  is  drawn  from  the  top  of  the  sign  over 

all  the  parts  of  it.  Thus  the  3d  root  of  47-f  22  is  -^47+22,  and 
the  2d  root  of  69  —  17  is  ^69—1 7,  &c. 

Sometimes  roots  are  designated  like  powers,  with  fractional  in- 

dices.  Thus,  the  squre  root  of  15, is  ]5'2,  the  cube  root  of  21  is  2P» 

and  4th  root  of  37  — r-  20  is  37— 20* ,  &c.  The  denominator  shows 
the  root  which  is  to  be  extracted,  and  the  numerator  shows  the 
power  to  which  that  root  is  to  be  raised.  Or  the  number  may  be 
raised  to  the  power  indicated  b}^  the  numerator,  and  the  root,  in- 

2.  2 

dicated  by  the  denominator,  then  extracted.     Thus  64"^=4  f=^lC. 

:?       3        3 
T=^Q4   =V^4096=16.     Hence  the  square  of  the  cube  root  of  any 

quantity  is  the  same  as  the  cube  root  of  the  square  of  the  same 
qjiantity. 

The  index  or  exponent  of  the  root  is  one  more  than  the  number 
of  muhiplications,  required  to  produce  the  given  number  orpow«?r. 


EVOLUTION". 


189 


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190  EXTRACTION  OF  THE  SQUARE  ROOTo 


THE  EXTRACTIOxX  OF  THE  SQUARE  ROOT. 

Rule. 

*1.  Distinguish  the  given  number  into  periods  of  two  figures 
each,  by  putting  a  point  over  the  place  of  units,  another  over  the 
place  of  hundreds,  and  so  on,  which  points  shew  the  number  of 
figures  the  root  will  consist  of, 

2.  Find  the  greatest  gquare  number  in  the  first,  or  left  hand 
period,  place  the  root  of  it  at  the  right  hand  of  the  given  number, 
(after  the  manner  of  a  quotient  in  division)  for  the  first  figure  of 
the  root,  and  the  square  number,  under  the  period,  and  subtract  it 
therefrom,  and  to  the  remainder  bring  down  the  next  period  for  a 
dividend. 

3.  Place  the  double  of  the  root,  already  found,  on  the  left  hand 
of  the  dividend  for  a  divisor. 

4.  Seek  how  often  the  divisor  is  contained  in  the  dividend,  (ex- 
cept the  right  hand  figure)  and  place  the  answer  in  the  root  for 
the  second  figure  of  it,  and  likewise  on  the  right  hand  of  the  divi- 
sor :  Multiply  the  divisor  with  the  figure  last  annexed  by  the  figure 
last  placed  in  the  root,  and  subtract  the  product  from  the  dividend  ; 
To  the  remainder  join  the  next  period  for  a  new  dividend. 

5.  Double  the  figures  already  found  in  the  root,  for  a  new  divi- 
sor, (or,  bring  down  your  last  divisor  for  a  new  one,  doubling  the 
light  hand  figure  of  it)  and  from  these,  find  the  next  figure  in  the 
root  as  last  directed,  and  continue  the  operation,  in  the  same  man- 
ner, till  you  have  brought  down  all  the  periods. 

Note  1.  If  when  the  given  power  is  pointed  off  as  the  power 
requires,  the  left  hand  figure  should  be  deficient,  it  must  neverthe- 
less stand  as  the  first  period. 

Note  2.  If  there  be  decimals  in  the  given  number,  it  must  b6 
pointed  both  ways  from  the  place  ot  units :  If,  when  there  are 

^  In  or;ler  to  slaew  the  reason  of  the  rale,  it  will  be  proper  to  premise  the 
following  Lemma.  The  product  of  any  two  numbers  can  have,  at  most,  but  so 
niaiiy  places  of  figures  as  ra-e  in  both  the  factors,  and  at  least  but  one  less. 

Dcvwnstraiion.  Take  two.r.umbers  consisting  of  any  number  of  places  ;  but 
let.  them  be  the  least  possible  of  those  placss,  viz.  Unity  with  cyphers,  as  100 
und  10  :  Then  their  product  will  be  1  with  so  many  cyphers  annexed  as  are  in. 
t)oth  the  aumbers,  viz.  1000 ;  but  1000  has  one  place  less  than  100  and  '  10  to- 
gether have  :  And  siace  100  and  10  wei-e  taken  the  least  possible,  the  product 
of  sny  other  two  numbers,  of  the  same  number  of  places,  will  be  greater  than 
li)00  ;  consequently,  the  product  of  any  two  numbers  can  have,  at  least,  but  one 
place  less  than  both  the  factors. 

Again,  take  two  numbers,  of  any  number  of  places,  which  shall  be  the  grcaJ- 
f  :t  possible  of  those  places,  as  03  and  9.  ■■  Now,  99  X  9  is  less  than  99  X  10  ;  but 
'JO  X  10  (:c:990)  contains  only  so  many  places  of  figures  as  are  in  99  and  9  ; 
therelore,  99  X  9,  or  the  product  of  any  other  two  numbers,  consisting  of  the 
-rme  number  of  places^  canaot  liave  Baore  places  of  figures,  than  are  in  both  its 
factors.  .'  -  \, 

('crollary  1;  A  square  uuiTibcr  cannot  have  more  places  of  figures  th%n,double 
♦•♦e  j-daoee  of  the'  root,  aiid  at  Ica^t  but  one  lesfe.  ; .'".  . 

^oroUci-ij  2.  A  cube  number  cannot  have  more  pla.ccs  of  figftrcs  Ihafi  triple  tlie 
rVv:rf  of  thsroo*.  rr'' ;-^  >    .*  i  ■  ^  ' ,    . -,   ,. 


EXTRACTION  OF  THE  SQUARE  ROOT.  191 

integers,  the  first  period  ia  the  decimals  he  deficient,  it  may  be 
completed  by  annexing  so  many  cyphers  as  the  power  requires  : 
And  the  root  must  be  made  to  consist  of  so  many  whole  numbers 
and  decimals  as  there  are  periods  belonging  to  each  ;  and  when 
the  periods  belonging  to  the  given  number  are  exhausted,  the  ope- 
ration may  be  continued  at  pleasure  by  annexing  cyphers. 

Examples. 
1st.  Required  the  square  root  of  30138696025  ? 

30138696026(173605  the  root. 

1 

1st.  Divisor=27)201 
189 

2d.  Divisor=343)1238 
1029 

3d.  Divisor=3466)20969 
20796 


4th.  Divisor=347205)  1736025 
1736025* 


I 


2d.  Required  the  square  root  of  5755  ? 

675'50(23'98+,  the  root, 
4 

43)175 
129 


469)4650 

4221 


4788)42900 
38304 


4596  Remainder. 

■^  The  Rule  for  the  extraction  of  the  square  root,  may  be  illustrated  by  at- 
tending to  the  process  by  which  any  number  is  raised  to  the  square.  The  seve- 
ral products  of  the  multiplication  are  to  be  kept  separate,  as  in  the  proof  of  the 
rule  for  Simple  Multiplication.     Let  37  be  the  number  to  be  raised  to  the  square, 

37  X  37= 1369====37  X  37 

37  37 

49=72  49=73 

21  .=3X7  >  _2x3x7  210=30X7  )  ^o^^Q^f 

21  .=3X7  S  — ^'^-^^  '  210=30X7  \ -^"^"^ 

9 .  .=32  <)00=302 


...     (37  (30-4-7=37 

fX3)42  .=2X3X7  "      * 

49^=72  rCarried  over. 


192  EXTRACTION  OF  THE  SQUARE  IlOOt. 

oti.  What  is  the  square  root  of  10342656  ?  Ans.  3216. 

•nil.   What  is  the  square  root  of  964-5192360241  ?  Ans.  31-05B7L  ' 
5th.  What  is  the  square  root  of  234  09  ?  Ans.  15-3. 

6th.  What  is  the  square  root  of -0000316969  ?  Ans.  -00563. 

7th  What  is  the  square  root  of  -046369  ?  Ans.  -213. 

Rules 
For  the  Square  Root  of  Vulgar  Fractions  and  Mixed  J^umlers. 

x\fter  reducing-  the  fraction  to  its  lowest  terms,  for  this  and  all 
other  roots  ;  then, 

1st.  Extract  the  root  of  the  numerator  for  a  new  numerator^  and 
the  root  of  the  denominator  iov  a  new  denominator,  which  is  the  best 
method,  provided  the  denominator  be  a  complete  power.  But  if 
it  be  not, 

2d.  Multiply  the  numerator  and  denominator  together  ;  and  the 
root  of  this  product  being-  made  the  numerator  to  the  denominator 
of  the  given  factor,  or  made  the  denominator  to  the  numerator  of 
it,  will  form  the  fractional  part  required.*     Or, 

Now,  it  is  evident  that  9,  in  the  place  of  hundredths,  is  the  greatest  square  iis 
this  product ;  put  its  root,  3,  in  the  quotient,  and  900  is  taken  from  the  product. 
The  next  products  are  21-f21=2x3x  7,  for  a  dividend.  Double  the  root  al- 
ready found,  and  it  is  2  X  3,  for  a  divisor,  which  gives  7  for  the  quotient,  whicli 
annexed  to  the  divisor,  and  the  whole  then  multiplied  by  it,  gives  2x3x7  (=42) 
-f-7x7  (=49)  which  placed  in  their  proper  places,  completely  exliausts  the-  re- 
mainder of  the  square.  The  same  may  be  shown  in  any  other  case,  and  the 
rule  become?  obvious. 

Perhaps  the  following  may  be  considered  more  simple  and  plain.     Let  37,=: 
30-f-7,  be  multiplied,  as  in  th^^monstration  of  simple  multiplication,  and  the 
products  kept  sei^arate, 
30+7 
30+7 


900+30x7 

30X7+49 


900+2x30x7+49=1369,  the  sum,  and  square. 
900  (30+7       The  root  of  900  is  30,  and  leaves 
the  two  other  terms,  which  are  ex- 


2X30+7x7)2x30x7+49  hausted  by  a  divisor,  formed  and 

2  X  30  X  7+49  multiplied  as  directed  in  the  rule. 


/7x'^         7 

'"^  Let  the  fraction  be  >/7^  then  by  the  rule,  >/l= — ^~~"" — "- =^l'o7+. 

•  -'  ^          --  v/7X2 

The  reason  of  which  is,  that  the  value  of  a  fraction  is  not  altered  by  multiplyino^ 

l)oth  its  parts  by  the  same  quajitity.    Thus  y/l~^  ^  ^      ^'     But  v'7X\/2= 

"        V7Xx/7_     7 

V7X2,  and  ^2Xv/2=2  evidently.     Ajid  thus  also,    ^^x^y~  n'-ycf 

/7  X2^ 

-and  is  the  rule.    Sco  Sards. 


EXTRACTION  OF  THE  SQUARE  ROOT.  193 

od.  Reduce  the  vulgar  fraction  to  a  decimal,  and  extract  its  root. 

4th.  Mixed  numbers  may  either  be  reduced  to  improper  fractions, 
and  extracted  by  the  first  or  second  rule,  or  the  vulgar  fraction 
may  be  reduced  to  a  decimal,  then  joined  to  the  integer,  and  the 
root  of  the  whole  extracted. 

Examples. 
1st.  What  is  the  square  root  of  y^flg  ? 
By  Rule  1. 

TiHo^TaaT  ^^C"*  *'°°^  of  the  numerator. 

16 

1681(41  root  of  the  denominator. 
16 


ol)81     Therefore,  -/y=the  root  of  the  given  fraction. 
81 

By  Rule  2. 
16x1681=26896,  and  ^26896=164.     Then, 

TVVT=TV4=TV=09756-f 

By  Rule  3. 

IG81)16(-0095181439H-.     And  ^/.0095181439=-097364-. 

2d.   What  is  the  square  root  of  f|f|  ?  Ans.  -f,^. 

3d.   What  is  the  square  root  of  42{  ?  Ans.  61-. 

Note.  In  extracting  the  square  or  cube  root  of  any  surd  num- 
ber, there  is  always  a  remainder  or  fraction  left,  when  the  root  is 
found.  To  find  the  value  of  which,  the  common  method  is,  to  an- 
nex pairs  of  cyphers  to  the  resolvend,  for  the  square,  and  ternaries 
of  C)'[)hers  to  that  of  the  cube,  which  makes  it  tedious  to  discover 
ihe  value  of  the  remainder,  especially  in  the  cube,  whereas  this 
trouble  might  be  saved  if  the  true  denominator  could  be  discovered. 

As  in  division  the  divisor  is  always  the  denominator  to  its  own 
fraction,  so  likewise  it  is  in  the  square  and  cube,  each  of  their  di- 
visors being  ihe  denominators  to  their  own  particular  fractions  or 
numerators. 

In  the  square  the  quotient  is  always  doubled  for  a  new  divisor  ; 
therefore,  when  the  work  is  completed,  the  root  doubled  is  the  true 
divisor  or  den(iminator  to  its  own  fraction  ;  as,  if  the  root  be  12, 
the  denominator  will  be  24,  to  be  placed  under  the  remainder, 
which  vulgar  fraction,  or  its  equivalent  decimal,  must  be  annexed 
to  the  quotient  or  root,  to  complete  it."'^ 

If  to  the  remainder,  either  of  the  square  or  cube,  cyphers  be  an- 
nexed, and  divided  by  their  respective  denominators,  the  quotient 
will  produce  the  decimals  belon,^insr  to  the  root. 

^*  Although  these  denominators  give  a  small  matter  too  much  in  the  square 
root,  and  too  little  in  the  cube,  yet  they  will  be  sufficient  in  common  ii?e.  i\n<l 
are  much  more  pxT>?iitioiT^  than  the  operation  -with  cvpher?. 

A  a 


191  APFLICATION  AND  USE 

APPLICATION  AND  USE  OF  THE  SQUARE  ROOT. 

Prob,  I.  To  find  a  7nean  proportional  between  two  numbers. 

Rule,  Blultiply  the  given  numbers  together,  and  extract  the 
square  root  of  the  product ;  which  root  will  be  the  mean  propor- 
tional sought. 

Note.     When  the  first  is  to  the  second  as  the  second  13  to  the 

third,  the  second  is  called  a  mean  proportional  between  the  other 

two.     Thus,  4  is  a  mean  proportional  between  2  and  8,  for  2  :  4  :; 

4X4 
4  ::  ~^=B,  or  4  is  as  much  greater  than  2,  as  8  is  greater  than 

4.  Bj' Theorem  I.  of  Geometrical  Proportion  2x8=4x4=4  .  To 
iind  a  mean  proportional  between  2  and  8,  take  the  square  root  of 
their  product.  The  same  must  be  true  in  every  case,  and  is  the 
rule. 

Example;. 
What  is  the  mean  proportional  between  24  and  96  ? 

V^96x24=48.     Answer. 
Prob.  II.  To  find  the  side  of  a  square  equal  in  area  to  arty  given 
superficies  whatever. 

Rule.  Find  the  area,  and  the  square  root  is  the  side  of  the 
Sijuare  sought.* 

Examples. 

1st.  If  the  area  of  a  circle  be  184125,  What  is  the  side  of  a 

square  equal  in  area  thereto  ?  

v/184  126=13-569-f  Answer. 

2d.  If  the  area  of  a  triangle  be  160,  what  is  the  side  of  a  square 
equal  in  area  thereto?  \/lG0=12*649-f-  Answer. 

Prob.  III.  A  certain  general  has  an  army  of  6625  men  :  pray 
How  many  must  he  place  in  rank  and  file,  to  form  them  into  a 
square  ?  V'5625=75  Answer.' 

Prob.  IV.  Let  10952  men  be  so  formed,  as  that  the  number  in 
rank  may  be  double  the  file.  74  in  file,  and  148  in  rank. 

Prob  V.  If  it  be  required  to  place  2016  men  so  as  that  there 
may  be  56  in  rank  and  36  in  file,  and  to  stand  4  feet  distance  in 
rank  and  as  much  in  file,  How  much  ground  do  they  stand  on? 

To  answer  this,  or  any  of  the  kind,  use  the  following  propor- 
tion :  As  unity  :  the  distance  ::  so  is  the  number  in  rank  less  by 
one  :  to  a  fourth  number  ;  next,  do  the  same  by  the  file,  and  mul- 

*  A  square  i.s  a  figure  of  four  equal  sides,  each  pair  meeting  perpentliculaily. 
or,  a  fl,^are  whose  length  and  breadth  are  equal.  As  the  area,  or  number  oi 
cquare  feet,  inches,  &c.  in  a  square,  is  equal  to  the  product  of  two  sides  whicli 
are  equal,  the  second  power  is  called  the  square.  Hence  tlie  rule  of  Prob.  II.  is 
evident. 

t  If  you  would  have  the  number  of  men  be  double,  triple,  or  quadruple,  &c. 
as  many  in  rank  as  in  file,  extract  the  square  root  of  ^,  i,  ^,  &;c.  of  the  given  num- 
ber of  men,  and  that  will  be  the  number  of  men  in  file,  whioli  double,  triplf. 
quadruple,  &c.  and  the  product  will  be  the  number  in  rank. 


I 


OF  THE  SQUARE  ROOJ,  195 

tiply  the  two  numbers  together,  found  by  the  above  proportion, 
and  the  product  will  be  the  answer.* 

As  1  :  4  ::  56~-l  :  220.  And,  as  1  :  4  ::  36—1  :  140.  Then. 
220X140=30800  square  feet,  the  Answer. 

Prob.  VI.  Suppose  I  would  set  ont  an  orchard  of  600  trees,  so 
that  the  length  shall  be  to  the  breadth  as  3  to  2,  and  the  distance 
of  each  tree,  one  from  the  other,  7  yards  :  Mow  many  trees  must 
it  be  in  length,  and  how  many  in  breadth?  and,  How  many  square 
yards  of  ground  do  they  stand  on  ? 

To  resolve  any  question  of  this  nature,  say,  as  the  ratio  in  length 
:  is  to  the  ratio  in  breadth  ::  so  is  the  number  of  trees  :  to  a  fourth 
number,  whose  square  root  is  the  number  in  breadth.  And  as  the 
ratio  in  breadth  :  is  to  the  ratio  in  length  ::  so  is  the  number  of 
trees  :  to  a  fourth,  whose  root  is  the  number  in  length. 

As  3  :  2  ::  600  :  400.     And  v^400=20=n umber  in  breadth. 

As  2  ;  3  ::  600  :  900.     And  v'900=30=nuraber  in  length. 

As  1  ;  7  ::  30—1  :  203.  And  as  1:7::  20—1  :  to  133.  And. 
203x133=26999  square  yards,  the  Answer. 

Prob.  VII.  Admit  a  leaden  pipe  ^  inch  diameter  tvill  fill  a 
cistern  in  3  hours;  I  demand  the  diameter  of  another  pipe  which 
will  fill  the  same  cistern  in  1  hour. 

Rule.  As  the  given  time  is  to  the  square  of  the  given  diameter, 
so  is  the  required  time  to  the  square  of  the  required,  diameter  j 
1=  75  :  and  •75x-75=  5625.     Then,  as  3h.  :  -5625  :: 
Ih.  :  1*6875  inversely,  and  \/l'6875=l  3, inches  nearly,  Ans, 

Prob  VHI.  If  a  pipe  whose  diameter  is  1  5  inches,  fill  a  cistern  iu 
5  hours,  in  what  time  will  a  pipe  whose  diameter,  is  3-5  inches  fill 
the  same  ? 

1-5x15=2  25;  and  3-5  x3'5=l 2-25.  Then,  as  2-25  :  5  ::  12'25 
:  918-f-hour,  inversely  =^55  min.  5  sec.  Answer. 

Prob.  IX.  If  a  pipe  6  inches  ])ore,  will  be  4  hours  in, running  off 
a  certain  quantity  of  water.  In  what  time  will  3  pipes,  each  four 
inches  bore,  be  in  discharging  double  the  quantity  ? 

6x6=36.  4X4=16,  and  16x3=48.  Then,  as  36  :  4h.  ::  48  :^ 
3h.  inversely,  and  as  Iw.  :  3h.  ::  2w.  :  6h.  Answer. 

Prob  X.  Given  the  diameter  of  a  circle,  to  make  another  circle 
which  shall  be  2,  3,  4,  &c,  tiroes  greater  or  less  than  the  given 
circle. 

Rur.E.  Square  the  given  diameter,  and  if  the  required  circle  bo 
greater,  multiply  the  square  of  the  diameter  by  the  given  pro- 
portion, and  the  square  root  of  the  product  will  be  the  required  di- 
ameter. But  if  the  required  circle  be  less,  divide  the  square  of 
the  diameter  by  the  given  proportion,  and  the  root  of  the  quotient 
will  be  the  diameter  required. 


*  The  abovo  rule  will  be  found  useful  ia  pUmting  trees,  having  the  distance  of 
ground  between  each  given. 

t  For  more  water  will  run  as  the  area  of  the  pipe  is  grcalei-,  nml  the  areas  of 
circular  pipes  vaj:y  as  the  square  of  tlxeir  diameter-. 


196  USE,  &c.  OF  THE  SQUARE  ROOT. 

There  is  a  circle  whose  diameter  is  4  inches  ;  I  demand  the  di" 
ameter  of  a  circle  3  times  as  large  ? 

4X4=16  ;  and  16x3=48;  and  ^48=6-928-1-  inches,  Answer. 

Prob.  XI.  To  find  the  diameter  of  a  circle  equal  in  area,  to  an 
ellipsis,  (or  oval)  whose  transverse  and  conjugate  diameters  are 
given.* 

Rule.  Multiply  th<»  two  diameters  of  the  ellipsis  together,  and 
the  square  root  of  that  product  will  be  the  diameter  of  a  circle 
equal  to  the  ellipsis. 

Let  the  transverse  diameter  of  an  ellipsis  be  48,  and  the  conju- 
gate 36  :  What  is  the  diameter  of  an  equal  circle  ? 

48x36=1728,  and  v'l 728=41. 569-{-the  Answer. 

Note.  The  square  of  the  hypolhenusfe,  or  the  longest  side  of  a 
right  angled  triangle,  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides;  and  consequently  the  difterence  of  the  squares  of 
the  hypothenuse  and  either  of  the  other  sides  is  the  square  of  the 
remaining  side. 

Prob.  XII.  A  line  36  yards  long  will  exactly  reach  from  the  top 
of  a  fort  to  the  opposite  bank  of  a  river,  known  to  be  24  yards 
broad.     The  height  of  the  wall  is  required? 

36x36  =  1296  ;  and  24x24=576.  Then,  1296—576=720,  and 
y'720=2e'J3-l-yards,  the  Answer. 

Prob.  XIII.  The  height  of  a  tree  growing  in  the  centre  of  a  cir- 
cular island  44  feet  in  diameter,  is  75  feet,  and  a  line  stretched 
from  the  top  of  it  over  to  the  hither  edge  of  the  water,  is  256  iept. 
What  is  the  breadth  of  the  stream,  provided  the  land  on  each  side 
of  the  water  be  level  ? 

256x256=65536:  and  75:^75=5625  :  Then,  65536— -5625= 
59911  and  V'59911=244-76-(-  and  244'76— ^=222-76  feet,  Ans. 

Prob.  XIV.  Suppose  a  ladder  60  feet  long  be  so  planted  as  to 
reach  a  window  37  feet  from  the  ground,  on  one  side  of  the  street, 
and  without  moving  it  at  the  foot,  will  reach  a  window  23  feet  high 
on  the  other  side  ;  I  demand  the  breadth  of  the  street  ? 

102.64  feet  the  Answer. 

Prob.  XV.  Two  ships  sail  from  the  same  port  ;  one  goes  due 
north  45  leagues,  and  the  other  due  west  76  leagues  :  How  far  are 
they  asunder?!  88*32  league*?,  Answer. 

Prob.  XVI.  Given  the  sum  of  two  numbers,  and  the  difierencr. 
of  \heir  squares,  to  find  those  numbers 

Rule.  Divide  the  diflerenr.e  of  their  squares  by  the  sum  of  the 
numbers,  and  the  quotient  will  be  their  diiierence.     The  two  num- 

*  The  tmnsverse  and  conjugate  are  the  longest  and  shortest  diameters  of  an 
<5llip!?is  ;  they  pass  throngh  the  centre,  and  cross  each  other  at  right  angles,  and 
the  diameter  of  the  equal  circle  is  the  square  root  of  the  product  of  the  diame- 
ters of  the  ellipsis. 

t  The  square  root  ma)'  in  the  same  manner  he  applied  to  navigation  ;  and, 
when  deprived  of  other  means  of  solving  problems  of  that  nature,  the  following 
proportion  will  serve  to  fuid  the  course. 

As  the  sura  of  the  hypothenuse  (or  distance)  and  half  the  greater  leg  (wheth- 
er difference  of  latitude  or  departure)  is  to  the  less  leg  ;  so  is  86,  to  the,  angle 
Ojjpositc  the  less  leg. 


EXTRACTION  OF  THE  CUBE  ROOT.      197 

bers  may  then  be  found,  from  their  sum  and  difference,  by  Prob. 
4,  page  57. 

Ex.  The  sum  of  two  numbers  is  32,  and  the  difference  of  their 
squares  is  256,  what  are  the  numbers  ? 

Ans.     The  greater  is  20.     The  less  12. 
Prob.  XVII.  Given  the  difference  of  two  numbers,  and  the  dif- 
ference of  their  squares,  to  find  the  numbers. 

Rule.  Divide  the  difference  of  the  squares  by  the  difference 
of  the  numbers,  and  the  quotient  will  be  their  sum.  Then  pro- 
ceed by  Prob.  4,  p.  57. 

Ex.  The  difference  of  two  numbers  is  20,  and  the  difference  of 
their  squares  is  2000  ;  what  are  the  numbers  ? 

Ans.     GO  the  greater.     40  the  less. 
Examples  for  the  two  preceding  problems. 

1.  A  and  B  played  at  marbles,  having  14  apiece  at  first  ;  B  hav- 
ing lost  some,  would  play  no  longer,  and  the  difference  of  the 
squares  of  the  numbers  which  each  then  had,  was  336  ;  pray  how 
many  did  B  lose  ?  Ans.  B  lost  6. 

2.  Said  Harry  to  Charles,  my  father  gave  me  12  apples  more 
than  he  gave  brother  Jack,  and  the  difference  of  the  squares  of 
our  separate  parcels  was  288  ;  Now,  tell  me  how  many  he  gave 
us,  and  you  shall  have  half  of  mine. 

Ans.     Harry's  share  12. 
Jack's  share      6. 

EXTRACTION  OF  THE  CUBE  ROOT. 

A  cube  is  any  number  multiplied  by  its  square.  To  extract  the 
cube  toot,  is  to  find  a  number  which,  being  multiplied  into  its 
square,  shall  produce  the  given  number, 

FIRST  METHOD. 

Rule. 
*1.  Separate,  the  given  number  into  periods  of  three  figures 
each,  by  putting  a  point  over  the  unit  figure,  and  every  third  fi- 
gure beyond  the  place  of  units. 

2.  Find  the  greatest  cube  in  the  left  hand  period,  and  put  its 
root  in  the  quotient. 

3.  Subtract  the  ctibe,  thus  found,  from  the  said  period,  and  to 
the  remainder  bring  down  the  next  period,  and  call  this  the  divi' 
dend. 

4  Alaltiply  the  square  of  the  quotient  by  300,  calling  it  the 
triple  square,  and  the  quotient  by  30,  calling  it  the  triple  quotient, 
and  the  sum  of  these  call  the  divisor. 

5.  Seek  how  often  the  divisor  may  be  had  in  the  dividend,  and 
place  the  result  in  the  quotient. 

*=  The  reason  of  pointing  the  given  number,  as  directed  in  the  rule,  is  obvious 
from  Coroll.  2,  to  the  Lemma  made  use  of  in  demonstrating  the  square  root. 

The  process  for  extracting  the  Cube  Root  may  be  iUustrated  in  the  same 
mauaer  as  that  for  the  Square  Root.     Jake  the  fame  number  37,  and  multiply 


}m  EXTRACTION  OF  THE  CUBE  ROOT. 

6.  Multiply  the  triple  square  by  the  last  quotient  figure,  and 
write  the  product  under  the  dividend  ;  multiply  the  square  of  the 
last  quotient  figure  by  the  triple  quotient,  and  place  this  product 
under  the  last ;  under  all,  set  the  cube  of  the  last  quotient  figure 
and  call  their  sum  the  subtrahend. 


as  before,  collecting  the  twice  21  into  one  sum,  as  they  belong  to 
ami  the  operation  will  be  simplified,  37  3=50653. 


the  same  place 


C    49=72 
:<  42-=2> 


49=72 
X3x7  420=2X30X' 


900=303 
7  the  multiplier.  37 


;73=^ 


f     343=73  343=7  3 

294 -=2X3X72  2940=2x30x7 

6:i  ■  -=33  X  7  0300=302  X  7 

1 47  -=3  X  7  2  1470=30  X  7  2 

1 20  •  -=2  X  32  X  7  1 2600=2  X  30  "  X ' 

27  •••=3^  27000=303 


27- ••  (37  '  27000   (30x7 

;X32M89* -=3x32x7  3 X 302 \1 8900=3 X 302  x' 
,X3   J  441—3x3x72      3x30/  4410=3X30X7 


As  27  or  27000  is  the  greatest  cube,  its  root  is  3  or  30,  and  that  part  of  tiie 
cube  is  exhausted  by  this  extraction.  Collect  those  terms  which  belong  to  the 
same  places,  and  we  have  32  x7=G3,  and  2x32  X 7=126,  and  63-1-126=3x32 
X7=189;  and2x3x72=294  and  3x72=147,  and  294-f  147=441=3X3X72, 
for  a  clixddend,  which  divided  by  the  divisor,  formed  according  to  the  rule,  the 
quotient  is  7,  lor  the  next  figure  in  the  root.  And  it  is  evident  on  inspecting 
the  work,  that  that  part  of  the  cube  not  exhausted  is  composed  of  the  several 
products  which  form  the  suhlrahend^  according  to  the  rule.  The  same  may 
be  shov/n  in  any  other  case,  and  the  universality  of  the  rule  hence  inferred. 

The  other  method  of  illustration,  employed  on  the  square  root,  is  equally 
applicable  in  Ihis'case. 

37:--;- 0-1- 7,  and  30-  .,         2x30x7-}-72 

30-|  7  the  multiplier. 


30"-f  2X302  x7-j- 30X72 

302x7-f2x30x72-}-73 

:.73:-50653=30;5-i-3x302x7-l-3x30x72 -1-7  3  (30-1-7=3'; 
30^ 


i;.vi:>  .^  3.C0O-     ..,,. ..J  :::x302  x7-f-3X 30x72-1-7 3  dividend. 

;:x302  X 74-3x50x73 -{-7 3  subtrahend. 

It  is  evident  that  303  is  the  greatest  cube.  When  its  root  is  extracted,  llie 
Bext  three  terms  eonslltute, the  dividend  ;  and,  tbe  several  products  formed  by 
means  of  the  quotient  or  second  figure  in  the  root,  are  precisely ^equal  to  the 
remainitig  parts  of  the  povier,  whose  root  was  to  be  found. 

The  arithmetical  demonstrations  of  the  Rules  for  extracting  either  the  square 
or  cube  root,  are  not  only  more  consistent  with  the  plan  of  an  Arithmctick  than 
demonstrations  on  the  figure,  called  a  square,  and  the  solid,  called  a  cube^  but 
they  are  much  more  readily  undereluo'i  1'}'  thoco  unaccustomed  lothe  mathcmati- 
«ulconside":  '^'i">  <i^  ^''^  i  br;  'inv. 


EXTRACTION  OF  THE  CUBE  ROOT.  159 

7.  Subtract  the  subtrahend  from  the  dividend,  and  to  the  re- 
mainder bring  down  the  next  period  for  a  new  dividend,  with 
which  proceed  as  before,  and  so  on  till  the  whole  be  finished. 

Note.  The  same  rule  must  be  observed  for  continuing  the  ope- 
ration, and  pointing  for  decimals,  as  in  the  square  root. 

Examples. 
a  St.  Required  the  cube  root  of  436036824287  ? 

436036824287(7583  root.7x 7x300=         14700=  1st. Trip,sq, 
343  7X     30     =  210=lst.do.quo, 


»t.  Din's —14910)93036=lst.  Dividend.  14910=  1st.  divisor- 


r3500  1 4700  X  5=         73500 

5250  5X5X210=  5250 

125  5X5X5     =  125 


78875=lst.  Subtrahend.  78875=l3t.  Subtra. 


2.Div.=1689750)  14161824=2d.  Divid.  75  X  75  X  300=    1687500=2d.  Trip.  ?q. 

75X30  =  2250=2d.  do.  quo. 

13500000  

144000  1689750=2d.  Divisor. 

512  — 

1687500X8=13500000 


1 36445 12=2d.  Subtra.  2250  X  8  X  8=      144000 
8X8X8       =  512 


^  Div.=172391940)517312287=3d.  Divid. 


517107600 


1 36445 12=2d.  Subtfa, 


204660  758  X  758  X  300=172369200=3d.  Trip.  sq. 

27  758X30  22740=3d.  do.  quo. 


5l7312287=3d.  Subtra.  17239 1940=3d.  Divisor. 


1 72369200  X  3=5 17 107600 
22740X3X3  =  204660 
3X3X3  =  27 


517312287=3d.  Subtra, 
2d.  What  is  the  cube  root  of  34965783?  Ans.  327. 

3d.  What  is  the  cube  root  of  84-604619?  Ans.  4-39. 

4th.  What  is  the  cube  root  of  008649  ?  Ans.  -2052+. 

5th.  What  is  the  cube  root  of  ^||?  Ans.  f. 

To  find  the  true  denominator^  to  he  placed  under  the  remainder,  after 
the  operation  is  ^nished. 
In  the  extraction  of  the  cube  root,  the  quotient  is  said  to  be 
squared  and  tripled  for  a  new  divisor ;  but  is  not  really  so,  till 
the  triple  number  of  the  quotient  be  added  to  it;  therefore  when 
the  operation  is  finis^hed,  it  is  but  squaring  the  quotient,  or  root, 
then  multiplying  it  by  3,  and  to  that  number  adding  the  triple  num- 
ber of  the  root,  when  it  will  become  the  divisor,  or  true  denomi- 
nator to  its  own  fraction,  which  fraction  must,  be  annexed  to  the 
quotient,  to  complete  the  root. 


200      EXTRACTION  OF  THE  CUBE  ROOT. 

Suppose  the  root  to  be  12,  when  squared  it  will  be  144,  and 
multiplied  b>'  3,  it  makes  432,  to  which  add  36,  the  triple  number 
of  the  root,  and  it  produces  468  for  a  denominator.* 

SECOND  METHOD. 

Rule. 

1.  Having  pointed  the  given  number  into  periods  of  three  figures 
iftacb,  find  the  greatest  cube  in  the  left  hand  period,  subtracting  it 
therefrom  and  placing'its  root  in  the  quotient ;  to  the  remainder 
bring  down  the  next  period  and  call  it  the  dividend. 

2.  Under  this  dividend  write  the  triple  square  of  the  root,  so 
that  units  in  the  latter  may  stand  under  the  place  of  hundreds  in 
the  former ;  and  under  the  said  triple  square,  write  the  triple  root, 
removed  one  place  to  the  right  hand,  and  call  the  sum  of  these  the 
divisor. 

3.  Seek  how  often  the  divisor  may  be  had  in  the  dividend,  ex- 
clusive of  the  place  of  units,  and  write  the  result  in  the  quotient. 

4.  Under  the  divisor  write  the  product  of  the  triple  square  of 
the  root  by  the  last  quotient  figure,  setting  down  the  unit's  place 
of  this  line,  under  the  place  of  tens  in  the  divisor  ;  under  this 
line,  write  the  product  of  the  triple  root  by  the  square  of  the  last 
quotient  figure,  so  as  to  be  removed  one  place  beyond  the  right 
hand  figure  of  the  former  ;  and,  under  this  line,  removed  one  place 
forward  to  the  right  hand,  writedown  the  cube  of  the  last  quotient 
figure,  and  call  their  sum  the  subtrahend. 

5.  Subtract  the  subtrahend  from  the  dividend,  and  to  the  re- 
mainder bring  down  the  next  period  for  a  new  dividend,  with  which 
proceed  as  before,  and  so  on  until  the  whole  be  finished. 

Example. 
Required  the  Cube  Root  of  16194277  ? 


*  It  may  not  be  amiss  to  remark  here,  that  t]ie  denominators,  botli  of  the 
square  and  cube,  shew  how  many  numbers  they  are  denominators  to,  that  is, 
what  numbers  are  contained  between  any  square  or  cube  number  and  the  next 
succcediir^  square  or  cube  number,  exclusive  of  botli  numbers,  for  a  complete 
number,  of  either,  leaves  no  fraction,  when  the  root  is  extracted,  and  conse- 
quently has  no  use  for  a  denominator,  but  all  the  numbers  contained  between 
them  have  occasion  for  it :  Suppose  the  square  root  to  be  12,  then  its  square  is 
144,  and  the  denominator  2^^,  which  will  be  a  denominator  to  all  the  succeeding 
numbers,  mitil  we  come  to  the  next  square  number,  vis.  169,  whose  root  is  lij, 
with  which  it  has  nothinjj  to  do,  for  between  the  square  numbers  144  and  169 
are  contained  24  numbers  excluding;-  both  the  square  numbers.  It  is  the  same 
in  the  cube  ;  for,  suppose  the  root  to  be  6,  the  cube  number  is  216,  and  its  de- 
nominator 126  will  be  a  denominator  to  all  the  succeeding  numbers,  until  avc 
.-•ome  to  the  next  cube  number,  viz.  343,  whose  root  is  7,  with  which  it  has  noth- 
ing to  do,  as  ceasing  then  to  be  a  denominator;  for  between  the  cube  343  and 
216  arc  126  numbers,  excluding  both  cubes.  And  so  it  is  with  all  other  denom- 
inatoi's,  cither  in  the  square  or  cube. 


EXTRACTION  O^  THE  CUBE  ROOT.  201 

16194277(253=Root. 
8 

8194         =  First  dividend. 


12  =  Triple  square  of  2. 

6  =a:  Triple  of  2. 

» 

126  =  First  divisor. 

60  ==  Triple  square  of  2,  multiplied  by  5. 

150  =  Triple  of  2  multiplied  by  the  square  of  5, 

125  =  Cube  of  5. 


7625         ==  First  Subtrahend. 


569277  =  Second  dividend. 


1875       =  Triple  square  of  25. 
75     ::=  Triple  of  25. 

18825     =  Second  divisor. 


B625       =  Triple  square  of  25  multiplied  by  3. 

675     =  Triple  of  25  multiplied  by  the  square  of  3. 
27  ==  Cube  of  3. 

569277  =^  Second  subtrahend. 

FIRST  METHOD  BY  APPROXIMATION. 
Rule. 

1.  Find,  by  trial,  a  cube  near  to  the  given  number,  and  call  it 
the  supposed  cube. 

2.  Then  as  twice  the  supposed  cube,  added  to  the  given  num- 
ber, is  to  twice  the  given  number,  added  to  the  supposed  cube, 
so  is  the  root  of  the  supposed  cube,  to  the  true  root,  or  an  ap- 
proximation to  it. 

3.  By  taking  the  cube  of  the  root,  thus  found,  for  the  supposed 
cube,  and  repeating  the  operation,  the  root  will  be  had  to  a  great- 
er degree  of  exactness. 

Example. 
It  is  required  to  find  the  cube  root  of  54854153  ? 
Let  64000000=supposed  cube,  whose  root  is  400  ; 
Then,  64000000         64854153 
2  2 

128000000        109708306 
54854153         64000000 
As  182854153    :    173708306"::  400 
400 

182854153)69483322400(379— root  nearly. 
B  b 


202  EXTRACTION  OF  THE  CUBE  ROOT. 

Again,  let  54439939  =  supposed  cube,  whose  root  is  370. 
Then,  54439939         54854153 
2  2 


108879878       109708306 
54854153  54439939 

As   163734031   :    164148246  ::  37J> 
379 


1477334205 
1149037715 
492444735 


l63734031)62212184855(379-958793+=rootcorFected. 

SECO?<D  METHOD  AY  APPROXIMATION. 
Rule. 

1.  Divide  the  resolvend  by  three  times  the  assumed  root,  and 
reserve  the  quotient. 

2.  Subtract  one  twelfth  part  of  the  square  of  the  assumed  root 
from  the  quotient. 

3.  Extract  the  square  root  of  the.  remainder. 

4.  To  this  root  add  one  half  of  the  assumed  root,  and  the  sum 
will  be  the  true  root,  or  an  approximation  to  it ;  take  this  approxi- 
mation as  the  assumed  root,  and,  by  repeating  the  process,  a  root 
farther  approximated  will  be  found,  which  operation  may  be  far- 
ther repeated,  as  often  as  necessary,  and  the  root  discovered  to 
any  assigned  exactness. 

Note.  In  order  to  find  the  value  of  the  first  assumed  root,  in 
this  or  any  other  power,  divide  the  resolvend  into  periods  by  be- 
ginning at  the  place  of  units,  and  including  in  each  period,  so  many 
figures  as  there  are  units  in  the  exponent  of  the  root;  viz.  3  figure* 
in  the  cube  root ;  4  for  the  biquadrate,  and  so  on  ;  then,  by  a  ta- 
ble of  powers,  or  otiierwise,  finrl  a  figure,  which  (being  involved 
to  the  power  whose  exponent  is  the  same  with  that  of  the  requir- 
ed root)  is  the  nearest  to  the  value  of  the  first  period  of  the  resol- 
vend at  the  left  hand,  and  to  that  figure  annex  so  many  cyphers  as 
there  are  periods  remaining  in  the  integral  part  of  the  resolvend; 
this  figure,  with  the  cyphers  annexed,  will  be  the  assumed  root, 
and  equal  to  r  in  the  theorem ;  and  it  is  of  no  importance 
whether  the  figure  thus  chosen  be,  when  involved,  greater  or  less 
than  the  left  hand  period,  as  the  theorem  is  the  same  in  both 
cases. 

1st.  What  is  Ihe  cube  root  of  1S6036824287  ? 


USE,  &c.  OF  THE  CUBE  ROOT.  9Md 

7000=assumed  root. 
3 


2 1  000)436036824287(20763658-2994 
Subtract  7000x7000^-12=4083333-3333 


-v/16680324-9661==4084'15^ 
Add  ^  the  assumed  root=3500 


\ 


And  it  gives  the  approximated  root=7584  15        ^ 
For  the  second  operation,  use  the  approximated  root  as  the  as- 
fiumed  (ine,  and  proceed  as  above. 

THIRD  METHOD  BY  APPROXIMATION. 

1.  Assume  the  root  in  the  usual  way,  then  multiply  the  square 
of  the  assumed  root,  by  3,  and  divide  the  resolvend  by  this  pro- 
duct ;  to  this  quotient  add  |  of  the  assumed  root,  and  the  sum  will 
be  the  true  root,  or  an  approximation  to  it. 

2.  For  each  succeeding  operation  let  the  last  approximated  root 
be  the  assumed  root,  and  proceeding  in  this  manner,  the  root  may 
be  extracted  to  any  assigned  exactness. 

1st.  What  is  the  cube  root  of  7  ? 

Let  the  assumed  root  be  2     Then.  2x2x3=12  the  divisor. 

12)7  0(  683  to  this  add  |  of  2=1  333,  &c.  that  is,  •683-{-l-333= 
1'916  approximated  root. 

Now  assume  1'916  for  the  root.     Then,  by  the  second  process, 

7 _^ 

the  root  is  — ii a+fx  1-916=1-9126,  &c. 

3x1-916 

2d.  What  is  the  cube  root  of  9  ?  Let  2  be  the  assumed  root  a6 
before.     Then,  j^^-|-|x  2=2  08  the  approximated  root    Now  as- 

9  

sume  2-08.     Then, :2-Hx2  08  =2-08008,  &c. 

3x208 

3d.  What  is  the  cube  root  of  282  ?  Let  6  be  the  assumed  root.— 
Then,  6x6x3=108)282(2  611,  &c.  and  2-61  l+f  of  6=6  611  ap- 
proximated root.  Now  assume  6  611,  and  it  will  be  6'611x6*61I 
X3=13M  16)282(2  1507,  &c.  and  2  1507 -f|  of  6-611=6-558  a 
farther  approximated  root- 

4th,  What  is  the  cube  root  of  1728  ? — Here  the  assumed  root  is 
tO-  Then,  10x10x3=300)1728(5-76,  and  5-76+|  of  10=12  426. 
—Now  assume  12426,  then  12  426  x  12  426x3=4"63-216428)1728 
(3-732,  and  3-732+§  of  12-426=12014  a  farther  approximated 
root,  and  so  on. 

APPLICATION  AND  USE  OF  THE  CUBE  ROOT. 

1.  To  find  two  mean  proportionals  between  any  two  given  num* 
bers. 

Rule. 

1.  Divide  the  greater  by  the  less,  and  extract  the  cube  root  of 
the  quotient. 


2^4  USE,  &c.  OF  THE  CUBE  ROOT. 

2.  Multiply  the  root,  so  found,  by  the  least  of  the  given  num- 
bers, and.  the  product  will  be  the  least. 

3.  Multiply  this  product  by  the  same  root,  and  it  will  give  the 
greatest.* 

Examples. 
1st.  What  are  the  two  mean  proportionals  between  6  and  760  ? 

-6=125,  and  ^  125=5.     Then,  6x6=30=Ieast,  and  30x 
greatest.     Answer  30  and  160. 
^oof.  As  6  :  30  ::  150  :  750. 
2d.   What  are  the  two  mean  proportionals  between  66  and  12096  ? 

Answer  336  and  2016. 
Note.  The  solid  contents  of  similar  figures  are  in  proportion  to 
each  other,  as  the  cubes  of  their  similar  sides  or  diameters. 

3d.  If  a  bullet  6  inches  diameter  weigh  32ife  ;  What  tvill  a  bul- 
let of  the  same  metal  weigh,  whose  diameter  is  3  inches  ? 

6x6x6=216.    3X3X3=27.     As  216  :  32ife  ::  27  :  41fe,  Ans. 
4th.  If  a  globe  of  silver  of  3  inches  diameter,  be  worth  £45, 
"What  is  the  value  of  another  globe,  of  a  foot  diameter  ? 

Ans.  £2880. 
The  side  of  a  cube  being  given,  to  find  the  side  of  that  cube 
which  shall  be  double,  triple,  &c.  in  quantity  to  the  given  cube.t 

Rule. 
Cube  your  given  side,  and  multiply  it  by  the  given  proportion 
isetween  the  given  and  required  cube,  and  the  cube  root  of  the 
product  will  be  the  side  sought. 

6th.  If  a  cube  of  silver,  whose  side  is  4  inches,  be  worth  £50,  I 
demand  the  side  of  a  cube  of  the  like  silver,  whose  value  shall  be 
4  times  as  much  ?  3 

4X4X4=64,  and  64x4=256.     V256=6-349-f  inches,  Ans. 
6th.   There  is  a  cubical  vessel,  whose  side  is  2  feet  ;  I  demand 
the  side  of  a  vessel,  which  shall  contain  three  times  as  much  ? 

Ans.  2ft.  lOf  inches. 
7th. I  The  diameter  of  a  bushel  measure  being  18|  inches,  and 
the  height  8  inches,  I  demand  the  side  of  a  cubic  box,  which  shall 
contain  that  quantity.  Ans.  12-9074-inches. 

*  As  two  mean  proportionals  are  required  to  two  given  numbers,  there  will 
he  four  terms  in  the  proportion,  in  which  the  first  is  to  the  second,  as  the  second 
to  the  third,  and  the  third  to  the  fourth.  The  numbers  therefore  belong  to 
a  geometrical  progression  of  four  terms.  The  first  part  of  the  rule  is  explained 
in  Prob.  VIII.  of  Geometrical  Progression,  and  the  second  and  third  parts  of  tlie 
fule  are  evident  from  the  proof  of  Prob.  I.  of  Geometrical  Progression. 

t  The  solid,  called  a  cube,  has  its  length  and  breadth  and  height  all  equal. 
As  the  number  of  solid  feet,  inches,  &c.  in  a  cube  are  found  by  multiplying  thej 
height  and  length  and  breadth  together,  that  is,  by  multiplying  one  side  into  it- 
self twice,  the  third  power  of  a  number  is  called  the  cube  of  that  number. 

I  Multiply  the  square  of  the  diameter  by  '7854,  and  the  product  by  the  height ; 
tiie  cube  root  of  the  last  product  is  the  answer.  Bee  Men  stir  atton  of  Svpcrjieif^ 
aiidSoMs.Art.^. 


EXTRACTION  OF  THE  BIQUADRATE  ROOT.        205 

S,  Suppose  a  ship  of  600  tons  has  89  feet  keel,  36  feet  beam,  and 
is  16  feet  deep  in  the  hold  :  What  are  the  dimensions  of  a  ship  of 
200  tons,  of  the  same  mould  and  shape  ? 

89x89x89=704969=cnbed  keel. 

As  500  :  200  ::  704969  :  281987-6  cube  of  the  required  keel. 

y'281987'6=65-57  feet  the  required  keel. 

As  89  :  65-57  ::  36  :  26  522=26^  feet,  beam,  nearly. 
As  89  :  65-57  :;  16  :  11-7  feet,  depth  of  the  hold,  nearly. 

9.  From  the  proof  of  any  cable  to  find  the  strength  of  any  other. 
Rule. — The  strength  of  cables,  and  consequently  the  weights 

of  their  anchors,  are  as  the  cubes  of  their  peripheries 

If  a  cable,  12  inches  about,  require  an  anchor  of  18cwt.  :  Of 
what  weight  must  an  anchor  be,  for  a  15  inch  cable  ? 

Cwt.  Cwt 

As  12X12X12  :  18  ::  15x15x15  :  35-15625  Ans, 

10.  If  a  15  inch  cable  require  an  anchor  35-15625cwt.  :  What 
must  the  circumference  of  a  cable  be,  for  an  anchor  of  18cwt? 

12  inches,  Answer. 

EXTRACTION  OF  THE  BIQUADRATE  ROOT. 

Rule. 
Extract  the  square  root  of  the  resolvend,  and  then  the  square 
root  of  that  root,  and  you  will  have  the  biquadrate  root. 
What  is  the  biquadrate  root  ol  20736  ? 

20736(144  144(12  root  required. 

i  1 


24)107  22)44 

96  44 

284)1136 
1136 

TWO  METHODS  OF  EXTRACTING  THE  BIQUADRATE  ROOT 
BY  APPROXIMATION. 


L 

^[  Rule 

1.  Divide  the  resolvend  by  six  times  the  square  of  the  assumed 
root,  and  from  the  quotient  subtract  j\  part  of  the  square  of  the 
assumed  root. 

2.  Extract  the  square  root  of  the  remainder. 

3.  Add  I  of  the  aesumed  root  to  ihe  square  root,  and  the  sum  will 
be  the  true  root,  or  an  approximation  to  it. 

4.  For  every  succeeding  operation,  either  in  this  or  the  follow- 
ing method,  proceeo  in  the  same  manner,  as  in  the  first,  each  lime 
using  the  last  approximated  root  for  the  assumed  root. 

The  biquadrate  root  of  20736  is  required. 
Here  10  is  the  assumed  root. 


2m         EXTRACTION  OF  THE  SURSOLID  ROOT. 

lOxlOx  6=600)20736(34-66 
Subtract  10x10-^18  =  6.6555 


V^290044=6-38 
Add  I  of  10     =6  63 

Approximated  root  1204,  to  be  made  the  as- 
sumed root  for  the  next  operation. 

Rule  1 1. 
Divide  the  resolvend  by  four  times  the  cube  of  the  assumed  root : 
to  the  quotient  add  three  fourths  of  the  assumed  root,  and  the  sum 
will  be  the  true  root,  or  an  approximation  to  it. 

Let  the  biquadrate  of  20736  be  required^  as  before  ? 
The  assumed  root  is  10 
10X10X10X4=::4000)20736(5.184 
Addfofl0=7-5 


Approximated  root  12-684,  ta  be  made  the  assumed 
root  for  the  next  operation. 

EXTRACTION  OF  THE  SURSOLID  ROOT  BY  APPROX- 
IMATION. 

A  Particular  Rule.* 

1.  Divide  the  resolvend  by  /Ive  limes  the  assumed  root,  and  to 
the  quotient  add  one  twentieth  part  o{  i\\Q  fourth  power  of  the  same 
root. 

2.  From  the  square  root  of  this  sum  subtract  one  fourth  part  of 
the  square  of  the  assumed  root. 

3.  To  tlie  square  root  of  the  remainder  add  one  half  of  ihe  assum- 
ed root,  and  the  sum  is  the  root  required,  or  an  approximation  to  it. 

Note.  This  rule  will  g^ive  the  root  true  to  five  places,  at  the  least, 
(an.d  generally  to  eight  or  nine  places)  at  the  first  process. 

Required  the  sur^olid  root  of  281950621875  ? 

200=assumed  root. 
5 

iouo"!     281950621-875  quotient. 

Add  200x200x200x200-r-20y  =80000000 

V^6 1950621 -875=  19025near]y. 
Subtract       200x  200-r-4=  10000 

^9025=95 
*  Add  half  the  assumed  root     =  100 

Required  root  195 


-^■r-'x-e-^-'/y/G    3:4      rr       r 
>    20     4       2 


I 


EXTRACTION  OF  THE  ROOTS.  207 

A  Oeneral  Rule  for  Extracting  the  Roots  of  all  Powers. 

1.*  Prepare  the  given  number,  for  extraction,  by  pointing  off 
from  the  unit's  place,  as  the  required  root  directs. 

2.  Find  the  first  figure  of  the  root  by  trial,  or  by  inspection  into 
the  table  Of  powers,  and  subtract  its  power  from  the  left  hand  pe- 
riod. 

3.  To  the  remainder  bring  down  the  first  figure  in  the  next  peri- 
od, and  call  it  the  dividend. 

4.  Involve  the  root  to  the  next  inferiour  power  to  that  which  is 
■given,  and  multiply  it  by  the  number  denoting  tiie  given  power, 
for  a  divisor. 

6.  Find  how  many  times  the  divisor  may  be  had  in  the  dividend, 
and  the  quotient  will  be  another  figure  of  the  root. 

6.  Involve  the  whole  root  to  the  given  power,  and  subtract  it 
from  the  giv€7i  number,  as  before. 

7.  Bring  down  the  first  figure  of  the  next  period  to  the  remain 
der  for  a  new  dividend,  to  which  find  a  new  divisor,  as  before,  aiul., 
in  like  manner,  proceed  till  the  whole  be  finished. 

Examples. 
1st.  What  is  the  cube  root  of  20346417  ? 

20346417(273         2X2x2=8  root  of  the  1st.  period,  or 

1st.  Subtrahend. 
23      =      8  z=i  1st.  Subtrah.2x2=4(=next  inferiour  power,)  and, 
—  4x3=(the  index  of  the  given  pow.)= 

12  Ist.  Divisor. 
22x3==:12)123=Dividend     27x27x27=  19683=2d.  Subtrahend. 

27x27=729  (next  inferior  power)  and, 

073    =       19683=2d.  Subt.729x3(=index  of  the   given  pow.)  = 

2187=2d.  Divisor. 
272x3=2I87)6634=2d.Di.273x273x273=27346417=3d.  Subtra. 

2733  =     20346417=3d.  Subtrahend, 


*  The  extracting  of  roo^s  of  very  high  powers  will,  by  this  rule,  be  a  tedious 
operatiori :  The  following  method,  when  practicable,  will  be  much  more  convoi" 
ient. 

When  the  index  of  the  power,  whose  root  is  to  be  extracted,  is  a  composite 
number,  take  any  two  or  more  indices,  Whose  product  is  equal  to  the  given  indez\, 
and  extract  out  of  the  given  number  a  root  answering  to  another  of  the  indices, 
and  so  on  to  the  last. 

Thus,  the  fourth  root=:square  root  of  the  square  root ;  the  sixth  root=:square 
root  of  the  cube  root;  the  eighth  root=:;square  root  of  the  fourth  root;  the 
ninth  root=the  cube  root  of  the  cube  root  ;  the  truth  root=;square  root  of  the 
fifth  root ;  the  twelfth  root=the  cube  root  of  tlie  fourth,  &c. 

The  general  rule  for  extracting  the  roots  of  all  powers,  may  be  illustrated  in 
the  same  way,  as  those  for  the  square  and  cube  roots.  Any  student  may  at  once 
see  the  truth  of  the  rule,  in  exhausting  the  several  products  of  the  case  illustrat- 
:ng  the  rule  for  the  cube  root.     And  thn  same  will  be  evident  by  raisiaa:  ihc 

iH'-li'^r  to  any  hi2:her  power. 


20^  EXTRACTION  OF  ROOTS 

Sd.  What  is  the  biquadrate  root  of  34827998976  ?  Ans.431-9'(' 
3d.  Extract  the  sursolid,  or  fifth  root  of  281950621 876? 

Ans.  195. 
4th.  Extract  the  square  cubed, or  sixth  root  of  1178420166016625? 

Ans.  326. 

A  General*  Rule  for  Extracting  Roots  by  Approximation. 

1.  Subtract  one  from  the  exponent  of  the  root  required,  and 

multiply  half  of  the  remainder  by  that  exponent,  and  this  product 

by  that  power  of  the  assumed  root,  whose  exponent  is  two  less  than 

that  of  the  root  required. 

*  The  general  theorem  for  the  extraction  of  all  roots,  by  approximation,  from 
whence  the  rule  was  taken,  and  the  Theorems  deducible  from  it,  as  high  as  tfee 
twelfth  power.  Let  G=:resolvend  whose  root  is  to  be  extracted.  p'-^e=  root 
required  ;  r  being  assumed  as  near  the  true  root,  and  w=:exponent  of  the  pow- 
cr-^then  the  equation  will  stand  thus. 

m — 2r^  m  — 2 

Hence, 


1 


G 

r    f 

wi — 2r^            m  — 2 

1                  ^ 

^V 

—    ...  ..  -^ r. 

m-1^  ^|m-.2 

2 

mm^ll"^          ^— i 

sm  for  the  cube  root     r-\-i 

y/G         rr         r 

3r         12         2 

For  the  Biquadrate    -     - 

y/G         rr         2r 
6»T      18         3 

For  the  Sursolid    -     -    - 

y/G        3rr       2r 
10r3    80          4 

For  the  squared  cube  root 

y/G         2rr        3r 
15r4    75          5 

For  the  second  sursolid    - 

^G         brr        5r 
21r5  252         6 

.       y/G         3rr        6r 

For  the  squared  Biquadrate 1 

28r6   196         7 

For  the  cubed  cube     -    - 

y/G         7rr        7r 
36r^  576        8 

For  the  squared  sursolid 

^/G         Arr       8r 
45r8  405         9 

For  the  third  sursolid     - 

v/G         9rr        Or 
55r9   1100     10 

y/G         5rr      10  r 

For  the  squared  square  cube    — 1 &c. 

66rio  726       11 

:j:  By  this  Theorem  the  fraction  is  obtained  in  numbers  to  the  lowest  terms  in 
all  the  odd  powers ;  and  in  the  even  powers  o»ly  by  having  the  numerator  and 
denominator  found  by  this  equation. 


BY  APPROXIMATION.  209 

2.  Divide  the  given  number  by  the  last  product ;  and  from  the 
quotient  subtract  a  traction,  whose  numerator  is  obtained  by  sub- 
tracting two  from  the  exponent,  and  multiplying  the  remainder  by 
the  square  of  the  assumed  root ;  and  whose  denominator  is  found 
by  subtracting  one  from  the  exponent  and  multiplying  the  square 
of  the  remainder  by  the  exponent. 

3.  After  this  subtraction  is  made,  extract  the  square  root  of  the 
remainder. 

4.  From  the  exponent  subtract  two,  and  place  the  remainder  as 
a  numerator  ;  then  subtract  one  from  the  exponent,  and  place  the 
remainder  under  the  numerator  for  a  denominator. 

5.  Multiply  this  fraction  by  the  assumed  root ;  add  the  product 
to  the  square  root,  before  found,  and  the  sum  will  be  the  root  re- 
quired, or  an  approximation  to  it. 

Example. 

What  is  the  square  cubed  root  of  1178420166015625  ? 

Let  the  assumed  root  =  300 

p     1 
Exponent  of  the  required  root  is  6.     Then,  x6=15. 

2 

3004=8100000000  and  this  multiplied  by  15=121500000000. 
1178420166015625-r-123  500000000=96989314,  from  this 

6—2x3002 

Subtract —  —2400 

6x6—12  


And  x/72y8y314=85-43 

6—2 

To  which  add  7^ — rX300=  240 

o — 1  ^ 

And  the  sum  is  the  approximated  root=         325^43 

For  the  2d  operation,  let  325-43  =  assumed  root. 

ANOTHER  METHOD  BY  APPROXIMATION." 

Rule. 

1.  Having  assumed  the  root  in  the  usual  way,  involve  it  to  that 
power  denoted  by  the.  exponent  less  1. 

*  A  rational  formula  for  extracting  the  root  of  any  pure  power  by  approxi- 
mation. 

Let  the  resolvend  be  called  G,  and  let  r-\-e  be  the  required  root,  r  being  as* 
sumed  in  the  usual  way. 

1  G  m~l 

Let  G  — be  required ;  then  r-\-e  = 1 —r  the  general  Theorem. 

in  m — 1  m 

mr 

G          2 
Hence,  For  the  cube  root  r-\-e  = 1 r. 

G  3 

For  th^  bi<iu?dratc     -     -  '   -[ -r. 

C  r. 


iiO 


SURDS. 


2.  Multiply  this  power  by  the  exponent. 

3.  Divide  the  resolvend  by  this  product,  and  reserve  the  quo- 
tient. 

4.  Divide  the  exponent  of  the  given  power,  less  1,  by  the  expo- 
nent, and  multiply  the  assumed  root  by  the  quotient. 

5.  Add  this  product  to  the  reserved  quotient,  and  the  sum  will 
be  the  true  root,  or  an  approximation. 

6.  For  every  succeeding  operation,  let  the  root  last  found,  be 
the  assumed  root. 

Example. 

What  is  the  square  cubed  root  of  1178420166016625  ? 
The  exponent  is  6.     Let  the  assumed  root  be  300. 
Then  3005x6=14580000000000 
1 4580000000000)  1 1 784201 6601 5625('80- 824. 
Add  I X  300=250 


330-824=approximated  root. 
For  the  next  operation,  let  330  824  be  the  assumed  root. 


SURDS. 

I.  SURDS  are  quantities,  whose  roots  cannot  be  obtained  exact- 
ly, but  may  be  approximated  to  any  definite  extent  by  continuing 
the  extraction  of  the  roots.  Surds  are  expressed  by  fractional  in- 
dices or  exponents,  or  by  the  radical  sign  V*     Thus.  3^,  or  \/3, 


G 

4 

For  the  sursoiiJ     -     - 

— 

+ 

— ?•. 

5r4 

5 

G 

5 

For  the  square  cubed 



+ 

— /•- 

6r5 

6 

G 

6 

For  the  seventh  root    - 

_ 

+ 

— r. 

7r6 

7 

G 

7 

For  the  eighth  -     -     - 

_ 

+ 

— r. 

8r7 

8 

G 

8 

For  the  ninth     -     -     - 

. 

+ 

— r. 

9r8 

9 

G 

9 

Pov  the  tenth     -     -     - 

-  _ 

H- 

— r. 

10/9 

10 

G 

10 

For  the  eleventli     -     - 

-b 

— r. 

llrio 

U 

G 

11 

For  the  twelfth    -     - 

. 

+ 

— r.  &c. 

V2rii 

12 

SURDS.  -211 

denotes  the  square  root  of  3.  The  value  of  2^  or  v^2,  to  the  hun- 
(Jreth  place  of  decimals,  is  1*41,  and  to  the  millionth  place  is 
1*41 1213.  The  value  of  s/9.  may  be  obtained  more  nearly  by 
continuing  the  extraction,  but  can  never  be  obtained  with  perfect 
accuracy,  as  is  easily  proved  in  the  following  section. 

Surds  are  often  called  irrational  quantities,  because  their  value 
cannot  be  expressed  by  figures.  They  are  thus  distinguished  from 
assignable  quantities,  which  are  called  rational  quantities.  Thus, 
2  is  a  rational^  and  ^2,  Slu  irrational  quantity. 

A  surd  is  always  connected  with  a  rational  quantity  expressed  or 
understood.  Thus,  as  the  square  root  of  2,  or  v'^,  is  that  quanti- 
ty taken  once^  unity  is  understood,  and  the  surd  is  expressed  either 
^/2,  lv/2,  or  lX\/2.  If  the  surd  is  to  be  taken  more  than  once, 
the  number  of  times  is  always  expressed  ;  thus  3-^/3,  or  3X\/3, 
means  thrice  the  square  root  of  3,  or  the  surd  taken  three  times. 

3 

Hence  it  is  that  an  expression  of  this  form,  l\/2,  or  3-v/5,  is 
considered  as  consisting  of  2  parts,  a  rational^  and  an  irrational  part, 
the  rational  part  always  expressing  the  number  of  times  the  surd 
is  taken. 

From  the  notation  of  powers,  and  surds,  these  expressions  arc 

equivalent;  viz.  3^=v^35  ;  and  23=^^22.  Also,  5-=v/5:^,  that 
is,  the  square  root  of  the  cube  of  5,  or  the  cube  of  the  square  root 
of  5. 

Note.  Though  surds  are  expressed  by  means  of  fractional  indi- 
ces or  the  radical  sign,  yet  it  is  common  to  apply  the  same  indices 
or  radical  sign,  to  complete  powers,  whose  roots  are  to  be  extract- 
ed. The  student  will  observe,  therefore,  that  quantities  expressed 
in  the  form  of  surds  are  not  necessarily  surd  quantities.  One  num- 
ber also  may  be  a  complete  power  of  on^  kind,  but  not  of  another. 
^  1.  1  1  J.  1 

Thus  4^  IS  2,  but  4'^  is  a  surd;  and  G4'-"  is  8,  and  G4Hs  4,  but  64* 

A  4  5 

and  64^  or  \/64  and  \/64  are  surds. 

II.  As  few  numbers  are  complete  powers,  surds  must  very  often 
occur  in  arithmetical  operations.  If  the  root  of  a  rs^hole  number  is 
not  a  whole  number,  neither  is  it  a  whole  number  and  a  decimal,  which 
can  be  assigned.  For,  supposing  the  entire  root  to  be  obtained, 
when  it  was  raised  to  the  power,  it  would  produce  a  whole  number 
and  a  decimal ;  while  the  supposition  requires  that  only  a  whole 
number  should  be  produced.  Thus,  supposing  the  square  root  of 
2,  or  \/2,  to  be  exactly  1-41,  or  1-414213,  this  root  raised  to  the 
square  should  produce  2  ;  but  it  is  obvious  that  the  square  would 
be  a  whole  number  and  a  decimal,  and  not  the  number  2. 

It  is  equally  evident,  that  the  root  of  a  vulgar  fraction  cannot 
be  assigned,  unless  both  parts  of  the  fraction  when  reduced  to  its 
lowest  terms,  are  complete  powers  of  the  roots  required.  Thus 
v'tg^VI— i;  but  y/^—^L  is  a  surd,  and  the  entire  value  of  the 
square  root  of  the  fraction  cannot  be  obtained. 


212  SURDS. 

III.  Though  the  value  of  a  surd  cannot  be  assigned,  its  power 

is  assignable.     From  the  definition  of  a  root,  it  is  evident  that  2- 
or  \/2  is  such  a  number  as  multiplied  by  itself,  the  product  or 

square  will  be  2.     Thus  \/2X\/2  or  22x2^=2.     And  S^xS^xS^^i 
3,  and  thus  for  other  surds. 

IV.  Arithmetical  calculations  are  often  simplified  by  certain  op- 
erations on  surds,  or  quantities  in  the  form  of  surds.  Rules  for 
several  operations  follow. 

1.  Any  number  may  be  reduced  to  the  foro^of  a  surd,  by  rais- 
ing it  to  the  power  denoted  by  the  index  of  the  surd,  and  then 
placing  the  power  under  the  radical  sign.  Thus  to  reduce  2  to 
the  form  of  the  square  root;  because  2x2=2^=4,  2=v'22=-v/4. 

s 
Reduce  2  to  the  form  of  the  fifth  root.  Ans.  v32. 

3 

lleduce  5  to  the  form  of  the  third  root.  Ans.  y  125. 

Reduce  7  to  the  form  of  the  fourth  root.  Ans.  ^'^^  /     . 

2.  Surds  are  reduced  to  their  most  simple  terms,  by  resolving  the 
quantity  under  the  radical  sign  into  two  factors,  one  of  which  shall 
be  a  complete  power  of  the  given  root ;  and  then  placing  the  root  of 
this  power  before  the  other  factor  under  the  radical  sign.     Thus 

v'27===-v/9x^'x/9X\/3==:3X\/3or3>s/3.  Aho, 1^32—1/16x^=1/ IS 

XV'2=2V'2. 

Reduce  \/l2o  to  its  most  simple  terms.         *  Ans.  5\/5. 

3  •  3 

Reduce  \/3384  to  its  most  simple  terms.  Ans.  8n/7. 

4 

Reduce  V^^j  to  its  simplest  terms.  Ans.  |Vf- 

3  7 

Reduce  \/481,  v/351,  and  \/896  to  their  most  simple  term?. 

Reduce  5\/20  to  its  simplest  terms.  Ans.  10\/5. 

Hence,  it  is  obvious,  that  if  a  factor  be  multiplied  into  a  surd, 
the  whole  may  be  reduced  tn  the  form  of  a  surd,  by  raising  the 
factor  to  the  power  denoted  by  the  surd,  multiplying  the  power  in- 
to the  surd,  and  placing  the  product  under  the  radical  sign.    Thus 

3v/3=^v/32xV3=:v79x3--=v'27  ;   and  8V7= V'8  ^xV 7= .7512x7== 

>/3584. 

3.  Surds  of  the  same  radical  sign  may  be  added  together,  when 
the  quantities  under  the  radical  sign  are  the  same,  by  prefixing  the 
sum  of  the  rational  parts  to  the  surd  quantity.     Thus  l\/24-l\/2, 

or  N/2-f  v/2=2x/2,  or  twice  V2  ;  and  oV5+W5~lV5. 

If  the  surds  are  not  already  in  their  most  simple  terms,  they  may 
often    be  added  after  the  reduction  is  made.     Thus  ^/20-|-v^80= 

2N/5-f-4v'5==6v'5  ;   and,  v/162-fV1350=3V2-f  5^2=8^2. 

What  is  the  sum  of  VBG  and  ^3584  ?  Ans.  10v^7. 

4.  Surds  of  the  same  radical  '•ign  may  be  subtracted,  if  the  surd 
part  he  the  game,  by  placing  the  difference  of  the  rational  parts 
before  tlie  $?:rd.     ff  the  quantities  are  not  already  in  their  simpierj-; 


SURDS.  213 

terms,  they  should  be  reduced  to  this  form.     Thus  n/4 — \/4=0  ; 
and  3x/3~-2v^3=lv/3  or  x/3.     Also  7^/5—4^5=3^5. 

4  4  4 

What  is  the  difference  between  v/1350  and  v/162  ?     Ans.  2v/2. 

J\''ote  1.  Surds,  apparently  incapable  of  addition  or  subtraction 
except  by  their  signs,  may  sometimes  be  reduced  to  a  commoa 
surd,  by  the  following  process,  and  their  sum  and  difference  readi- 
ly found.  Thus  let  the  surds  be  \/|  and  \/f  As  v/|=v/|xl=^ 
Vi=^2Vh  and  as  V^=V^\=Wi^  then  ^i+v^f =2Vi+fVi==: 
Wi=W^\='2XW^=TW^*  their  sum:  And  2Vi-iVi^Wi=' 
jjVQy  their  difference. 

What  is  the  sum  and  difference  of  \/|  and  f\/|  ? 

Ans.  Their  sum  is  ^VQ.     Their  diff.  is  -^\VS. 

What  is  the  sum  of  ^^^v/lS  and  \/^^^  ?  Ans.  V|. 

What  is  the  difference  of  v^f  and  n//^  ?  Ans.  ^v^lo. 

JVote  2.  If  the  same  quantity  is  under  different  radical  signs,  or 
if  the  same  radical  sign  has  different  quantities  under  it  when  the 
surds  are  in  their  simplest  terms,  the  surds  can  be  added  or  sub- 
tracted only  by  the  signs  of  addition  or  subtraction.     Thus  it  is 

3 

evident,  that  y/2-\-V2,  is  neither  twice  the  square   root  of  2  nor 

3 

twice  the  cube  root  of  2  ;  and  that  3\/3— 2\/3,  is  neither  the  square 
root  of  3  nor  the  cube  root  of  3.  It  is  equally  obvious,  that  2y'3 
-|-2\/2,  is  neither  four  times  the  square  of  3  nor  of  2  ;  and  that  4v/2 
— 2v/3,  is  neither  twice  the  square  of  2  nor  of  3. 

5.  Surds  of  the  same  radical  sign  are  multiplied  like  other  num- 
bers, but  the  product  must  be  placed  under  the  same  radical  sign, 

Thu|  v/27Xn/64  =  V'27x64=n/1728=12,   for   v'27=3,   and   ^04 

=4,   and   ^27x^64=3x4=12.      And   V'2x  V3=V"^xS  =  VG. 

Also  3v/3x4n/5=12V15  or  v^27X\/80,  and  V27X\/80=x/27x80 
=  12n/15=v/2160.  • 

Sometimes  the  product  of  the  surds  becomes  a  complete  power 
of  that  root,  and  the  root  should  then  be  extracted,  as  in  the  first 
of  the  preceding  examples.  Also  in  this  example  ;  v'2Xv'200= 
v/400=20. 

It  is  evident  from  the  first  example  in  this  section,  that,  when 
quantities  are  under  the  same  radical  sign,  the  root  of  the  product 
of  qvantities  is  equal  to  the  product  of  their  roots. 

If  a  surd  be  raised  to  a  power  denoted  by  the  index  of  the  root, 

the  power  will  be  rational.  Thus,  \/3X\/3,  or  32x3-'^=3.  In 
this  example  2  is  the  index  of  the  root,  and  the  surd  is  raised  to 

3  3  3 

the  second  power  or  square.  Also  \/4X\/4xv'4=4.  If  fractional 
indices  be  used,  in  order  to  multiply  surds  of  the  same  root,  you 

l_        JL        X  _3 

have  only  to  add  the  indices.  Thus  43x43x4^=43=4'  or  4,  uni- 
ty being  the  implied  index  of  4,  or  of  the  first  power  of  any  num- 
ber.    In  all  cases  when  the  sum  of  the  numerators  contains  the  com- 


^'14 


bUKDS. 


7non  denominator  a  certain  number  of  times  exactly ^  the  product  tt'u'/ 
be  rational.     Thus  3^x3^x3^=3      3      =3^=32=8,   and  72x7'-- 

-:r7-T~=7  -  =75. 

As  5"*  may  be  expressed   according  to  the   notation  ot'  powers, 

4  3  4.  ii  4±'i  8 

thus,  6*,and  53  by  5',  hence  5^X5'=5  ]'=5'=o».  Therefore, 
to  multiply  different  powers  of  the  same  root,  you  have  only  to  add 
the  indices  of  the  i^iven  root,  and  place  the  sum  for  the  index  oi' 
tlie  power  which  is  produced.  Thus  32x32=3'*,  or  the  square  of 
a   number  multiplied  by  itself  produces  the  fourth   power;   the 

cube  by  the  cube,  the  sixth  power,  and  so  on.     Thus  also  2*x2^  — 
5_L.  y        8        2 

6.  Surds  of  the  same  radical  sign  are  divided  like  whole  num- 
bers, but  the  quotient  must  be  placed  under  the  same  radical  sign. 

Thus    -/1728--^^64=vi^--S34=V2'7==3;    and    V^-^V^— 

^t)-^3=\/2.  Sometimes  the  quotient  becomes  a  complete  power, 
as  in  the  lir>t  example,  in  which  case  the  root  should  be  extracted. 

So  also  in  the  following;    ^400—^  100=  v'^i^^-^^^t)=  V4=2. 

As  Vl'J'28  — 12,  and  ^64— 4,  then  ^  1720-^v'64=12~4=3=^ 
3 
^27.     Hence,  the  quotient  of  the  roots  of  quantities  is  the  same  as 

the  root  of  their 'quotient ^  if  the  quantities  are  under  the  same  radi- 
cal sign. 

Divide    ^  \0^    by  >^6.     Xow  V  108-4-VG=  V^08^^=  V  18  = 

Now  9^^100=^8100,  and  3^2  = 


V9x2=3-v^2. 

Divide   9^100  by  3^2. 
V  18,  and  v/8i00H-v^l8=v/810Q~-18=v/450=15v/2.     Or  Qv/lOO 

-^3v^^=9-T-3x  V 400-^^2  =  9~-3x  ^  100-r-2=9H-3  X  a/ bQ=^ 
3v/50=Sx5v/2=15v/2. 

Divide  -^V'yW  ^v  ^Vf  Now-}-f-^=|;  and  \^ ii^-^ V 1'=  \^  hS 

^f v''f  =^^^^2  ;  and  |xiv/2=fv/2. 

Divide  ^"48  by  v'-.V ;  W60  by  3^15  ;  and  |>/-^  by  iv'!- 
If  the  quantities  under  the  radical  sign  be  the  same,  the  quotient 
will  be  found  by  dividing  the  rational  parts  only.     Thus  \/2~x/2 

3  3 

=v/li=l,  or  5/2  is  contaliied  in  v/2  once.     Also  iv/3+iv/3=2, 
and  2v/5+5\/5=3^. 

To  divide  one  power  by  another  of  the  same  root,  place  the  dif- 
ference of  the  indices  for  the  index  of  the  given  root.  'J'his  is  mere- 
ly reversing  a  process  given  in  t)je  preceding  section.  The  rea- 
son of  the  process  may  aljjO  be  seen  in  the  following  manner.    Thu.^ 

2**         2^X22  3  8  3 

^■".--C!^:^; — =r. — r=:2- =2'-'3^     A!^o.  22-!-2*  =2''-;"2* ,  ]>v  reduc 


SURDS.  215 

4 


ji-9 


|_v/2^ 


ing  the  indices  to  a  common  denominator ;    and  2* 

4|2^         4    r^^X^S  4  4 5. 

If  the  index  of  the  divisor  exceed  that  of  the  dividend,  the  index 
of  the  quotient  will  be  their  difference  with  the  sign  of  subtraction 

before  it.      Thus,  5--r-5^=5--'^=5-^.      Now,  as  52-t-3s=— = 

52         ^      1  1 

——-=—,   5-5=—.,     Hence  a  power  whose  index  has  the  sign 

of  subtraction  before  it,  is  the  same  powerof  the  reciprocal  of  that 
quantity.  Hence,  there  is  an  obvious  method  of  transferring  pow- 
ers from  the  numerator  to  the  denominator  of  a  fraction.  Thus, 
1  3-3         1  2 

2^=2-2^  and  -7-=.  ^23'  and  ^37=2x3-.  There  is  also  an  obvious 

method  of  finding  the  value  of  a  quantity  whose  index  has  the  sign 

of  subtraction  before  it.    Thus  2-3=-~=_  and  7[~''=T[~''=  l^- 

11  _i      1        1       v/2 

orr25-=:^^=.^5^=16.     And  2  ^=-7=;^-=— 

7.  To  raise  surds  to  any  power,  multiply  the  index  of  the  surd 
by  the  index  of  the  required  power.  Thus  2^  raised  to  the  square 
is  22  +  2=23=^22^^=2  ;  and  the  cube  of  3*  =3*"^ *'''*=  3"*''^'''' 

If  there  be  rational  parts  with  the  surd,  they  must  be  raised  to 
the  given  power,  and  prefixed  to  the  required  power  of  the  surd. 

Thus,  3v/3,  or  3X3^   raised  to  the  square,  is  3=  X3'^^2=32x3^ 

=32X33^^=9n/32,  or  9x/9.     And  the  cube  of  22=2^'^3+i=r 

22X3=2^=>/23=v/8=:2V'2,  when  reduced  to  its  simplest  terms. 
Also,  the  fourth  powerof  Ax/2  is  2i(r^22=^^-. 

Required  the  fifth  power  of  Ix/j;. 

8.  To  extract  any  root  of  a  surd  quantity,  divide  the  index  of 
the  quantity  by  the  index  of  the  required  root.     Thus,  the  square 

1^  _l_::-0  1.  6  ?.  3-:-'s 

root  of  23  is  23  *  "==2«  or  \/2,  and  the  cube  root  of  3*  is'  3*  ' ' 

3 
--=3^x4=34  Q.  ^3. 

If  there  be  rational  parts  with  the  surd,  their  root  must  be  pre- 
fixed to  the  required  root  of  the  irrational  part.     Thus,  the  square 

root  of  9x/9,  or  9x93=92v/32=3v/3.  The  process  must  evident- 
ly be  the  reverse  of  that  in  the  preceding  section,  and  the  reason 
of  it  is  obvious. 


216  S'URDb. 

What  is  the  cube  root  of-^V^S  ?  Ans.  |. 

What  is  the  square  root  of  10*  ?  Ans.  lO^  or  100^/10. 


Examples. 

-■  "f  JL  2  0 

1.  MuUiplj  G*  by  G%  and  the  product  is  G-^^  or  v/G^. 

2.  Divide  6"^  by  G%  and  the  quotient  i§  6^^  or  -v/6. 

3  3  3 

:3.  Add  \/32  and  \/108,  and  multiply  the  same  by  x/yf^. 

Ans    IG^orgv/S, 

3  3  3 

4.  Addv32  and  \/100,  and  divide  (he  sum  by  \/,4j-  Ans.  5^. 

5,  Find  the  shortest  method  of  dividing  3  by  v/2,  to  any  given 
place  of  decimal?. 

3  3Xx/2       3v/2     x/18     4-242640  &c 

^"^'^  ~v/7=;7iW2="^"^-*^= i =2  121320 

&c. 


6.  Find  tlie  sum  of  x/'j  and  x/aV'  ^"^  ^^^^  their  difference. 

3  3  ;{  3 

Ans.  Their  surii  is  1^/54,  or  f >/2.     Their  diff.  is  |v/2,  or  \/^. 

3  3 

7.  What  is  the  sum  and  difference  of  n/|  and  \//^. 

Ans.  Their  sum  is  /^v/lB.     Their  diff.  ^\s/lQ. 

8.  There  are  four  spheres  each  4  inches  in  diameter,  lying  so 
as  to  touch  each  ether  in  the  form  of  a  square,  and  on  the  middle 
of  this  square  is  put  a  fifth  ball  of  the  same  diameter ;  what  is 
the  perpendicular  distance  between  the  two  horizontal  planes 
which  pass  through  the  centres  of  the  balls? 

4  4v/2 

Ans.  -^-=— ^=2v/2=:V'8=2'8284-f  inches; 

JVote,  It  may  be  seen  from  this  example  that  the  diameter  of  the 
ball  divided  by  v^2,  will  give  the  distance  between  the  planes, 
whatever  be  the  diameter  of  the  ball,  or,  which  is  the  same,  half 
the  diameter  of  the  ball  multiplied  by  the  square  root  of  2. 

9.  I'here  are  two  balls,  each  four  inches  in  diameter,  which 
touch  each  other,  and  another,  of  the  same  dinmetcr  is  so  placed 
])etween  them  that  their  centres  are  in  the  same  vertical  plane  ; 
what  is  the  distance  between  the  horizontal  planes  which  pass 
through  their  centres  ? 

Ans.     r— =|V3=2V'3  inches. 

jYoic.  It  is  evident  from  this  example,  that  in  all  similar  cases, 
half  the  diameter  of  the  ball  multiplied  by  the  square  root  of  3, 
gives  the  distance  between  the  planes. 

10.  There  is  a  quantity  to  whose  square  i  is  to  be  added  ;  of 
the  sum  the  square  root  is  to  be  taken  and  raised  to  the  cube  ;  to 
this  power  -*-  are  to  be  added,  and  the  sum  will  be  yVlS  ;  what  i« 
thai  quantity  ?  Ang.  VL 


PROPORTION.  217 

OF  PROPORTION  IN  GENERAL. 

NUMBERS  are  compared  together  to  discover  the  relations 
they  have  to  each  other. 

There  must  be  two  numbers  to  form  a  comparison  :  the  number, 
which  is  compared,  being  written  first,  is  called  the  ajitecedent ;  and 
that,  to  which  it  is  compared,  the  consequent. 

Numbers  are  compared  with  each  other  two  different  ways :  The 
one  comparison  considers  the  dijjference  of  the  two  numbers,  and  is 
called  arithmetical  relation,  the  difference  being  sometimes  named 
the  arithmetical  ratio  ;  and  the  other  considers  their  quotient^  which 
is  termed  geometrical  relation,  and  the  quotient,  the  geometrical 
ratio.     Thus,  of  the  numbers  12  and  4,  the  difference  or  arith- 

12 
metical  ratio  is  12 — 4=8  ;  and  the  geonxetrical  ratio  is  -7-  =3,  and 

of  2  to  3  is  \. 

\(  two,  or  more,  couplets  of  numbers  have  equal  ratio'?,  or  dif- 
ferences, the  equality  is  termed  proportion  ;  and  their  terms,  simi- 
larly posited,  that  is,  either  all  the  greater,  or  all  the  less  taken 
as  antecedents,  and  the  rest  as  consequents,  are  called  proportion- 
als. So  the  two  couplets  2.  4,  and  G,  8,  taken  thus,  2,  4,  6,  8,  or 
thus,  4,  2,  8,  G,  are  arithmetical  proportionals;  and  the  two  coup- 
lets, 2,  4,  and  8,  16,  taken  thus  2,  4,  8,  16,  or  thus,  4,  2,  IG,  8,  are 
geometrical  proportionals.* 

*  To  denote  numbers  as  being-  geometrically  proportional,  the  couplets  are 
separated  by  a  double  colon,  and  a  colon  is  written  between  the  terms  of  each 
couplet;  we  may,  also,  denote  arithmetical  proportionals  by  separating  the 
couplets  by  a  double  colon,  and  writing  a  colon  turned  horizontally  between  the 
terms  of  each  couplet.  So  the  above  arithmeticals  may  be  written  thus,  2.-4 
::  6  ..  8,  and  4  ..  "2  ::  «  .-  6  ;  where  the  first  antecedent  is  less  or  greater  than  its 
consequent  by  just  so  much  as  the  second  antecedent  is  less  or  greater  than  it;* 
consequent :  And  the  geometricals  tlius,  2  :  4  ::  8  :  16,  and  4  :•  2  ::  16:8;  where 
the  first  antecedent  is  contained  in,  or  contains  its  consequent,  just  So  often,  as 
the  second  is  contained  in,  or  contains  its  consequent. 

Four  numbers  are  said  to  be  reriprocalh/  or  inversely  proportional,  when  the 
fourth  is  less  than  the  second,  by  as  many  times,  as  the  third  is  greater  than  the 
first,  or  when  the  first  is  to  the  third,  as  the  fourth  to  the  second,  and  vice  versa. 
Thus  2,  9,  6  and  3,  are  reciprocal  proportionals. 

Note.  It  is  common  to  read  the  geometricals  2  :  4  ::  8  :  16,  thus,  2  is  to  4  as  8 
to  16,  or.  As  2  to  4  so  is  8  to  16. 

Harmonical  proportion  is  that,  which  is  between  those  numbers  which  assign 
the  lengths  of  musical  intervals,  or  the  lengths  of  strings  sounding  musical  notes  ; 
and  of  tkree  numbers  it  is,  when  the  Jirst  is  to  the  third,  as  the  difference  belweeu 
the  first  and  second  is  to  the  difference  between  the  second  and  third.,  as  the  num- 
bers 3,  4,  6.  Thus,  if  the  lengths  of  strings  be  as  these  numbers,  they  will  sound 
an  octave  3  to  6,  a  fiftli  2  to  3,  and  a  fourth  3  to  4. 

Again,  between  4  numbers,  when  the  Jir.it  is  to  the  fourth,  as  the  difference  be- 
tween the  first  and  second  is  to  the  difference  between  the  third  and  fourth,  as  in  the 
numbers  5,  6,  8,  10  ;  for  strings  of  sych  lengths  will  sound  an  octave  5  to  10 ;  u 
.sixth  greater,  6  to  10  ;  a  third  greater  8  to  10  ;  a  third  less  5  to  6  ;  a  sixth  less 
5  to  8  ;  and  a  fourth  6  to  8. 

Let  10,  12,  and  15,  be  three  numbers  in  harmonical  proportion,  then  by  the 
precpim°;  defiftition.  10  •.  15  ::  12 — 10  :  15—12,  and  bvTlicorem  I.  of  Geometri-n 

P   d 


iiii>  i^KOPORTlON. 

Proportion  is  tlistinguisheil  into  continued  and  discontinued,  if. 
of  several  couplets  of  proportionals,  written  down  in  a  series,  the 
difference  or  ratio  of  each  consequent,  and  the  antecedent  of  the 
next  following  couplet,  be  the  i^ame  as  the  common  difference  or 
ratio  of  the  couplets,  the  proportion  is  said  to  be  continued,  and 
the  numbers  themselves,  a  series  of  continued  arithmetical  or  ge- 


cal  Proportion,  lOx  lo— iii=15X  l:ii— 10,  or  10X15— 10x12— 15X12— 15X 
10,  whence  if  any  two  of  the  three  terms  be  given,  the  other  may  be  found  ia 
tlie  following  manner. 

Case  1.  Given  the  1st  and  2d  terms  to  find  the  3d. 

As  lOX  15— IPX  12— 15X 12— 15X  10,  then  jO X 15- 15X  12-}-15X  10=:10X 
12,  or  2X10X15—12X15=10X12,   or,  "^iu— 12  X  15=10  X 1 2,  and   15= 

10X12 

—.  that  is,  15,  the  ihird  is  equal  to  the  product  of  the  first  and  second 

2X10—12 

ttrms^  divided  by  the  difference  of  twice  the  first  term  and  the  second  term. 

2.  Given  the  1st  and  3d  to  find  the  second  term. 

From  the  same  equivalent  expression,  we  get  2X  lOX  15=15X  12-f  lOX  12= 

j_ 2X10X15 

15-|- 10X12, and — — — :--,  =^12,  that  is,  the  5gconrf  term  is  equal  to  twice  the 

lU— |-  lo 
prod^ict  of  the  first  and  third  terms^  divided  by  the  sum  of  the  first  and  second 
terms. 

3.  Given  the  second  and  third  to  find  the  fii-st  term. 

From  the  same  expression,  we  get2XlOXl5 — 10X12=15X12,  or  2X15—12 
15X12 
X  10=15X12,  and  10  ^ — -  _  ■■-— ,  that  is^  the  first  term  is  equal  to  the  product 

of  the  second  and  third  terms^  divided  by  the  difference  of  twice  the  third  term 
and  the  second  term. 

Ex.  Find  from  tliird  term,  or  monochord,  50,  and  the  first  term,  or  octave^  25, 
the  second  term. 

2  X  25  X  50     2500 

By  Case  2,  — — '—=^—^ — =33'33,  the  second  term,  and  is  the  length  cif 

25+50      .75 
that  chord,  which  is  called  a  fifth. 

If  there  be  four  harmonical  proportionals,  as,  5,  6r  B  and  10  ;  then,  according 

to  the  definition,  5  :  10 ::  6—3  :  10—8,  and  as  before,  5  X  10— 8=10X0— 5,  or  5X 
10 — 5X0=10X6^ — 10X5.  From  this  expression,  we  may  find  any  one  of  four 
harmonical  proportionals  from  the  other  three.     Thus,  the  first  three  being  giv 

5X8 

en  to  find  the  fourth;  2x10X5— 10  XG=5X  8,  and  10= --,  that  is,  the 

2Xi> — 6 
fourth  term  is  equal  to  the  product  of  the  first  and  third  divided  _  by  the  difference 
of  twice  the  first  teryn  and  the  second  term. 

In  the  same  manner,  it  may  be  shown,  that  the  third  term  of  four  harmonical 
proportionals  is  equal  to  the  difference  of  tic  ice  the  product  of  the  first  andfouvth 
terms  and  t/te  product  of  the  second  and  fnirih  terms.,  divided  by  the  first  term. 

''^X5Xl0 6x10 

If  the  terms  be  5,  6,  8,  and  10,  then  8=^^—-^ ■ . 

5 
Also,  The  second  term  is  eqiial  to  the  difference  of  twice  the  fourth  and  tJif 
third  temiy  multiplied  *>y  the  quotient  of  the  fir  at  divided  by  the  fourth  term,     h 

the  terms  be  as  before,  C=2x  10 — 8x— • 

Also,  The  .//«/ term  is  equal  to  the  product  of  the  second  and  fourth  terms, 
divided  by  the  diffcrancc  of  twice  the  fourth  and  the  third  frrvi.     Thw^  5— 

cxio 
f^xio— t;" 


ARITHMETICAL  PROPORTION.  219 

jinetiical  proportionals.  So  2,  4,  6,  8,  form  an  arilhmelical  pro- 
gression ;  for  4 — 2=6 — 4=8 — 6=2  ;  and  2,  4,  8, 16,  a  geometric 
cal  progression  ;  for  |=f=V^=^2. 

But,  if  the  jlifference  or  ratio  of  the  consequent  of  one  couplet, 
and  the  antecedent  of  the  next  couplet  be  not  the  same  as  the  com- 
mon difference  or  ratio  of  the  couplets,  the  proportion  is  said  to 
be  discontinued.  So  4,  2,  8,  6,  are  in  discontinued  arithmetical 
proportion  ;  for  4—2=8 — 6=2=rommon  difference  of  the  coup- 
lets, 8 — 2=:6=difference  of  the  consequent  of  one  couplet  and 
the  antecedent  of  the  next;  also,  4,  2,  16,  8,  are  in  discontinued 

4     16   ■ 
geometrical  proportion  ;  for  ~=-g-=^2=common  ratio  of  the  coup- 

16  .     ' 

lets,  and  -^=8=ratio  of  the  consequent  of  one  couplet  and  the 

antecedent  of  the  next. 


ARITHMETICAL  PROPORTION. 

Theorem  I. 

IF  any  four  quantities  2,  4,  6,  8  be  in  arithmetical  proportion,* 
the  sum  of  the  two  means  is  equal  to  the  sum  of  the  two  ex- 
tremes.! 

And  if  any  three  quantities,  2,  4,  6,  be  in  arithmetical  propor- 
tion, the  double  of  the  mean  is  equal  to  the  sum  of  the  extremes. 

Theorem  II. 

In  any  continued  Arithmetical  Proportion  (1,  3,  5,  7,  9,  11)  the 
sum  of  the  two  extremes,  and  that  of  every  other  two  term.^, 
equally  distant  from  them,  are  equal.     Thus,  1  -{-1  l=3-{-9=5-|-7.J 

When  the  number  of  terms  is  odd,  as  in  the  proportion  3.  8.  13. 
18.  23,  then,  the  sum  of  the  tW4>  extremes  being  double  to  the 
mean  or  middle  term,  the  sum  of  any  other  two  terms,  equally 
remote  from  the  extremes,  must  likewise  be  double  to  the  mean, 

*  Althouj^h  in  the  comparison  of  quantities  according  to  their  diiTerences,  the 
term  proportion  is  used  :  yet  the  word  progression.,  is  frequently  .substituted  in 
its  room,  and  is  indeed  more  proper  ;  the  former  form  beinj]^,  in  the  common  ac- 
ceptation of  it,  synonymous  with  ratio,  which  is  only  used  in  the  other  kind  of 
comparison. 

t  For  since  4 — 2=8—6,  therefore  4-f  6=2+8. 

ij:  Since,  by  the  nature  of  jjrogressionals,  the  second  term  exceeds  the  first  by 
just  so  much  as  its  corresponding;  term,  the  last  but  one,  wants  of  the  last,  it  is 
evident  that  when  these  corresponding  terms  are  added,  the  excess  of  the  one 
will  make  good  the  defect  of  the  other,  and  so  their  sum  be  exactly  the  same 
w;th  that  of  the  two  extremes,  and  in  the  same  manner  it  will  appear  that  the 
sum  of  any  two  other  corresponding  torm?  must  be  equal  to  that  of  the  two 
extreme". 


220  ARITHMETICAL  PROGRESSION. 

Theorem  III. 
In  any  continued  Arithmetical    Proportion,  as  4,  44.2,  44-4, 
44-6,  4-|-8,&c.  the  last  or  greatest  term  is  equal  to  the  sum  of  the 
first  or  least  term  and  the  common  difference  of  the  terms,  multi- 
plied by  the  number  of  the  terms  less  one.* 
Theorem  iV. 
The  sum  of  any  rank,  or  series  of  quantities  in  continued  Arith- 
metical Proportion  (1.  3.  5.  7.  9.  11.)  is  equal  to  the  sum  of  the 
two  extremes  multiplied  into  half  the  number  of  terms.t 


ARITHxMETICAL  PROGRESSION. 

ANY  rank  of  numbers,  more  .than  two,  increasing  by  a  common 
excess,  or  decreasing  by  a  common  difference,  is  said  to  be  in 
Arithmetical  Progression. 

If  the  succeeding  terms  of  a  progression  exceed  each  other,  it 
isc?ll"d  an  ascending  series  or  progression  ;  if  the  contrary,  a  de- 
scending series. 
o         ^  U.  2.  4.  6.     8.  10,  &c.  is  an  ascending  arithmetical  series. 

}  1.  2.  4.  8.  16.  32,  &c.  is  an  ascending  geometrical  series. 
.    ,    i  10.    8.  6.  4.  2.  0,  &c.  is  a  descending  arithmetical  series. 

(  32.  16.  8.  4.  2.  1,  &c.  is  a  descending  geometrical  series. 

*  For  since  each  term,  after  the  first,  exceeds  that  preceding  it  by  the  com- 
mon difference,  it  is  plain  that  the  last  must  exceed  the  first  by  so  many  times 
the  common  difference  as  there  are  terms  after  the  first ;  and  therefore  must  be 
equal  to  the  first,  and  the  common  difference  repeated  that  number  of  times. 

t  For,  because  (by  the  second  Theorem)  the  sum  of  the  two  extremes,  and 
that  of  every  other  two  terms,  equally  remote  from  them  are  equal,  the  whole 
Feries,  consisting  of  half  so  many  such  equal  sums  as  there  are  terms,  will  there- 
fore be  equal  to  tlie  sum  of  the  two  extremes,  repeated  half  as  many  times  as 
there  are  terms. 

The  same  thing  also  holds,  when  the  number  of  terms  is  odd,  ns  in  the  scric? 
A^  8,  12,  16,  20  ;  for  then,  the  mean,  or  middle  term,  being  equal  to  half  the  sura 
of  any  two  terms,  equally  distant  from  it  on  contrary  sides,  it  is  obvious  that  the 
value  of  the  whole  series  is  the  same  as  if  every  term  thereof  were  equal  to  the 
mean,  and  therefore  is  equal  to  the  mean  (or  half  the  sum  of  the  two  extremes) 
multiplied  by  the  whole  number  of  terms  ;  or  to  the  sum  of  the  extremes  mul- 
tij^lied  by  half  the  number  of  terms. 

'['he  sum  of  any  number  of  terms  of  the  aritlimetical  series  of  odd  numbers^ 
1,  3,  5,  7,  9,  fec.is  equal  to  the  square  of  that  number. 
For,  0+1  or  the  sum  of  1  term   =  12  or    1 
1+3  or  the  sum  of  2  terms  =  22  or    4 
4+5  or  the  sum  of  3  terms  =  32  or    9 
9+7  or  the  sum  of  4  terms  =  4?  or  IG 
16+9  or  the  sura  of  f>  terms  =:  .53  or  2.5,  &c. 
By  continuing  the  addition,  tlie  rule  would  be  true  for  any  number  of  term?. 

EXAMPI,K. 

']']ie  fa-st  term,  the  ratio,  and  number  of  terms  given,  to  find  the  sum  of  the 
series, 

A  g:entleman  travelled  29  days,  the  first  day  he  went  but  1  mile,  and  increased 
every  day's  travel  2  miios  ;  How  far  did  he  travel  ?      29X29^=^41  miles,  Ans. 


ARITHMETICAL  PROGRESSION.  221 

The  numbers  which  form  the  series,  are  called  the  terms  of  the 
progression. 

Note,  The  first  and  last  terms  of  a  progression  are  called  the 
extremes,  and  the  other  terms  the  means. 

Any  three  of  the  five  following  things  being  given,  the  other 
two  may  he  easily  found. 

1.  The  first  term. 

2.  The  last  term. 

3.  The  number  of  terms. 

4.  The  common  difference. 

5.  The  sum  of  all  the  terms.  ^ 

Problem  F. 

The  first  ierm^  the  last  term,  and  the  number  of  terms  beifig  given,  (o 
find  the  common  difference. 

Rule.* 
Divide  the  difference  of  the  extremes  by  the  number  of  terms 
less  1,  and  the  quotient  will  be  the  comtnon  difference  sought. 

Examples. 
1st.  The  extremes  are  3  and  39,  and  the  number  of  terms  is  19  : 
What  is  the  common  difference  ? 

Extremes. 


39  > 
-3^ 


Divide  by  the  number  of  terms  less  1  =  19-— 1  =  18)36(2  Ans, 

36 
39—3  — 

Or,  =2. 

19—1 
2d.  A  man  had  10  sons,  whose  several  ages  differed  alike  ;  the 
youngest  was  3  years  old,  and  the  eldest  48  :  What  was  the  com- 
mon difference  of  their  ages?  Ans.  5. 

3d.  A  man  is  to  travel  from  Boston  to  a  certain  place  in  9  days, 
and  to  go  but  5  miles  the  first  day,  increasing  every  day  by  an  equal 
excess,  so  that  the  last  day's  journey  may  be  37  miles :  Required 
the  daily  increase  ?  Ans.  4. 

Problem   IF. 

The  first  iermy  the  last  term,  and  the  number  of  terms  being  given,  to 

find  the  sum  of  all  the  terms. 

RuLE.f  Multiply  the  sum  of  the  extremes  by  the  number  of 
terms,  and  half  the  product  will  be  the  answer. 

*  The  difference  of  the  first  and  last  terms  evidently  shews  the  incrrase  of 
the  first  terra  by  all  the  subsequent  additions,  till  it  becomes  equal  to  the  last ; 
and  as  the  number  of  those  additions  was  one  less  than  the  number  of  terms,  and 
the  increase,  by  every  addition,  equal,  it  is  plain  that  the  total  increase,  divided 
by  the  number  of  additions,  must  give  the  difference  of  every  one  separately  j 
whence  the  rule  is  manifest. 

t  Suppose  another  series  of  the  same  kind  with  the  given  one  be  placed  under 
/  in  an  inverse  order  ;  then  will  the  sum  of  any  two  corresponding  terms  be  the 


222J  ARITHMETICAL  PROGRESSION. 

Examples. 

1st.  The  extremes  of  an  arithmetical  series  are  3  and  39^  aitd 
the  number  of  terms  19  :  Required  the  sum  of  the  series  ? 


,    g  >  Extremes. 


Sum=42 
Number  of  terms  =  x   1 9 

378 

42 

2)798 


39-f3xl9 

Or, =399.  399  Ans. 

2 
2J.  It  is  required  to  find  how  many  strokes   the  liammer*'of  a 
dock  would  strike  in  a  week,  or  168  hours,  provided  it  increased 
1  at  each  hour?  Ans.  14196. 

3<i.  Suppose  a  number  of  stones  were  laid  a  j'ard  distant  from 
each  other  for  the  space  of  a  mile,  and  the  first  a  jard  from  a  bas- 
ket :  What  length  of  ground  will  that  man  travel  over,  who  gath- 
ers them  op  singly,  returning  with  them  one  by  one  to  the  basket? 

Ans.  1761  miles. 

N.  B.  In  this  question,  there  being  1760  yards  in  a  mile,  and  the 
snan  returning  with  each  stone  to  the  basket,  his  travel  will  be 
doubled ;  therefore  the  first  term  will  be  2,  and  the  last  1760x2> 
and  the  number  of  terms  1760. 

4tb.  A  man  bought  25  yards  of  linen  in  Arithmetical  Progress- 
ion ;  for  the  4(h  yard  he  gave  12  cents,  and  for   the  last  yard  75 
cents  :  What  did  the  whole  amount  to,  and  what  did  it  average 
per  yard  ? 
75—12 

— =3  the  common  difference  hy  which  the  first  term  is  found 

22—  1  [to  be  3. 

75+3x25 

Then — =J9  75b.  and  the  average  price  is39cts.  per  yard 

2 
6th.  Required  the  sum  of  the  first  1000  numbers  in  ther  natural 
order?  Ans.  500500. 


dme  as  that  of  the  first  and  last ;  consequently,  any  one  of  those  sums,  multiplied 
.'y  the  number  of  terms,  must  give  the  whole  sum  of  the  two  series. 
Let     1,  2,  3,  4,  5,  6,  7,  8,  be  the  given  series. 
And   8,  7,  6,  5,  4,  3,  2,  1,  the  same  inverted. 
Then,  9+9+9+9-f  9+9+9+9=9  X  8=72,  and 

72 
1+2+3-f4+5-fC+7-fO=— =36. 


ARITHMETICAL  PROGRESSIO^J:.      '  223 

Problem  III. 

Given  the  extremes  and  the  common  difference^  to  find  the  namh^r  of 

terms. 

Rule.*  Divide  Ihe  difference  of  the  extremes  by  the  com- 
mon difference,  and  the  quotient  increased  by  1  will  be  the  num- 
ber of  terms  required. 

Examples. 

1st.  The  extremes  are  3  and  39,  and  the  common  difference  2: 
What  is  the  number  of  terms  ? 

on   i 

Extremes. 


39? 
-  3^ 


Common  difference =2)36 

Q.uotient=18 

Add    1 

3g__3  

Or,- h  1^19.  19  Ans. 

2 
2d..  A  man  going  a  journey,  travelled  the  first  day  7  miles,  tUc 
last  day  51  miles,  and  each  day  increased  his  journey  by  4  miles : 
How  many  days  did  he  travel,  and  bow  far  ? 

Ans.  !2  days^  and  348  miles. 

Problem  IV*. 

TTie  extremes  and  common  difference  given^  to  find  the  sum  of  all 

the  series. 

Rule.  Multiply  the  sum  of  the  extremes  by  their  difference  in- 
creased by  the  common  difference,  and  the  product  divided  by 
twice  the  common  difference  will  give  the  sum.j 

Examples. 
1st.  If  the  extremes  are  3  and  39,  and  the  common  difference  2  : 
What  is  the  sum  of  the  series  ? 


*  By  the  first  Problem,  the  difference  of  the  extremes,  divided  by  the  number 
of  terms  less  1,  gave  the  common  difference  ;  consequently  the  same  divided  by 
the  common  diflerence,  must  give  the  number  of  terms  less  1 ;  hence,  this  quo- 
tient, augmented  by  1,  must  be  the  answer  to  the  questicm. 

t  By  the  3d  Problem  find  the  number  of  terms,  and  then,  with  the  number  of 
terms  and  the  extremes,  find,  by  Prob.  2,  the  sum  of  the  series.     This  is  the  rule, 

39 3 

v/hich  is  contracted  in  the  text.     Thus  in  the  1st  Example,  by  Problem  3, 


|-l=rthe  number  of  terms,  and  by  Prob.  2,  39-f-3X39 — 3-j-l=twice  the  sum  of 


■D  ^  30— T         .      ,      39^3_i-2                ^        3y+3Xri^'— ^H-!iI 
!;ie  series.     But  —- 1-1  is  also  ~—   Therefore.  — — Tv"; '" 

'i-^  ?iim  of  the  eerie?,  and  i«  the  rule. 


224  ARITHMETICAL  PROGRESSION. 

39-1-3=42  sum  of  the  extremes. 
39 — 3=='36=diiference  of  extremes. 

364-2=;38=difference  of  extremes  increased  by  the  coo>moL 
difference. 

42 
X    38 

336 
126 

Twice  the  common  difference=4)1596 

399 


39+3x39— 3-{-2 

Or, =399. 

2X2 
2d.  A  owes  B  a  certain  sum,  to  be  discharged  in  a  year,  by  pay- 
ing 6d.  the  first  week,  18d.  the  second,  and  thus  to  increase  every 
weekly  payment  by  a  shilling,  till  the  last  payment  be  21.  1  Is.  6d. : 
What  is  the  debt  ?  Ans.  £67  12:.. 

Pkoblem  V. 
The  extremes  and  sum  of  the  series  given,  to  find  the  number  of  tenm 

Rule. 
Twice  the  sum  of  the  series,  divided  by  the  Sum  of  the  extremes, 
will  give  the  number  of  terms.* 

Examples. 
1st.  Let  the  extremes  be  3  and  39,  and  the  sum  of  the  series 
399  :     What  is  the  number  of  terms  ? 

Sum  of  the  series=399 
X     2 

Sum  of  the  extremes=39+3=42)798(19  Ans. 

42 

.378 
399x2  378 

Or, =19. 

39+3 

2d.  A  owes  B  671.  12s.  to  be  paid  weekly  in  Arithmetical  Pro- 
ofression,  the  first  payment  to  be  6d.  and  the  last  to  be  51s.  6d. : 
How  many  payments  will  there  be,  and  how  long  will  he  be  in  dis- 
charging the  debt  ? 

Ans.  52  payments,  and  as  many  weeks. 

*  This  Problem  is  the  reverse  of  Prob.  II.  and  the  reason  of  the  rule  is  ob^  i 
ous  from  tlie  demonstration  of  the  Rule,  Prob.  II, 


ARITHMETICAL  PROGRESSION.  22^, 

Problem  VI* 

Che  extremes  and  the  sum  of  the  series  given,  to  Jind  the  common 

difference. 

Rule. 
Divide  the  product  of  the  sum  and  difference  of  the  extremes, 
by   the  difference  of  twice  the  sum  of  the  series,  and  the  sum  of 
the  extremes,  and  the  quotient  will  be  the  common  difference.* 

Examples. 
1st.  Let  the  extremes  be  3  and  39,  and  the  sum  399 :    What  is 
the  common  difference  ? 

Sum  of  the  extremes  =  39  +  3  =     42 
Diff.  of  the  extremes  =  39—3  =  X  36 

252 
126 


399X2—42=756)1512(2  Anav 
1512 


3y_j-3x  39—3 

Or,  _--     .-=2. 

399x2—39-1-3 
2d.  A  owes  BJ£67   12s.  to  be  discharged  in  a  year,  by  weekly 
ipayments  ;  the  first  payment  to  be  6d.  and  the  last,  JC2   lis  6d.  ; 
What  is  the  common  difference  of  the  payments,  and  what  will  each 
payment  he  ? 

61*5-1-  5x51*5 — '5 

—  .         —  .=18.  and  6d.4'l3.=  ls.  6d.=2d  payment, 

1352x2—51-54-5 
Is.  6d.-{-ls.=2s,  6d.=3d  payment,  &c. 

Problem  VII. 

The  first  term,  the  common  difference,  and  the  number  of  terms  given, 
to  jind  the  last  term. 

Rule. 
The  number  of  terms  less  1,  multiplied  by  the  common  differ- 
ence, and  the  first  term  added  to  the  product,  will  give  the  last 
term.t 

Examples. 
1st.  If  the  first  term  be  3,  the  common  difference  2,  and  the 
number  of  terms  19  :   What  is  the  last  term  ? 

*  This  rale  is  only  a  contraction  of  the  following  process.  By  Prob.  V.  find 
the  number  of  terms,  and,  then,  from  the  extremes  and  number  of  terms,  find  by 
JProb.  I.  the  common  difference. 

t  By  Prob.  I.  the  diffcreixee  of  the  last  and  first  terms  divided  by  the  number 
of  terms  less  1,  gives  tjie  common  difierence,  whence  the  common  ditference 
multiplied  by  the  number  of  terms  less  1,  and  the  product  increased  by  the  first, 
term,  must  give  the  last  term. 

K  e 


:?26  ARITHMETICAL  PROGRESSION. 

Number  of  terms  —  19 
—  1 


Number  of  terras  less  1  =  18 
Common  difference  =  X  2 


36 
First  term  =  +  3 


39  the  Ans. 


Or,  19— lXi-^+3=r39. 

2d.  A  owes  B  a  certain  sum  to  be  paid  in  Arithmetical  Prop:rei3- 
slon  ;  the  first  payment  is  6d.  the  number  of  payments  52,  and  the 
common  difference  of  the  payments  is  12d.  :  What  is  the  last  pay- 
ment?  Ans.  £2  lis.  6d. 

Problem  Vllf. 

Thejirsi  term,  common  diff'erence^  and  number  of  terms  gmen^  to  find 
the  $uni  of  the  series. 

Rule. 

To  the  first  term  add  the  product  of  the  number  of  terms  less  I 
by  half  the  common  difference,  and  their  sum,  multipHed  by  the 
number  of  terms,  will  give  the  sum  of  the  progression.* 

Examples. 
1st.  If  the  first  term  be  3,  the  common  difference  2,  and  number 
of  terms  19  :  What  is  the  sum  of  the  series  ? 

First  term  =  3 

Add  the  product  of  the  number  of  terms  }  _.iq iw| in 

less  1  by  ^  common  difference  5 

Their  sum  21 
Multiply  by  the  number  of  terms  =  li' 

189 
21 


Or,  19x3+19—1x1=399  Ans.  =  399 

2d.  Sixteen  persons  gave  charity  to  a  poor  man  ;  the  first  gave^ 
7c.  and  the  second  12c.  and  so  on  in  arithmetical  progression  ;  I 
demand  what  sum  the  last  person  gave,  and  how  much  the  poor 
man  received  in  all  ? 

Ans.  82c.  the  last  gave,  and  ^7  12c.  the  whole  sura. 

*  Find  by  Prob.  VII.  the  last  term,  and  then  by  Prob.  II.  Uxe  sum  of  the  pro  - 
§*ression.     Tl*e  rule  is  merely  a  contraction  of  this  process. 


ARITHMETICAL  PROGRESSION.  1^27 

Problem  IX. 
Given  the  first  term^  number  of  terms,  and  the  sum  of  the  serieSy  to 
find  the  last  term. 
Rule. 
Divide  twice  the  sum  by  the  number  of  terras  ;  from  the  quo- 
tient take  the  first  term,  and  the  remainder  will  be  the  last.* 

Examples. 
1st.  If  the  first  term  be  3,  the  number  of  terms  19,  and  the  sum 
399  ;  What  is  the  last  term  ? 

Sum  of  the  terms  ==  399 
Multiply  by       2 


I)ivide  by  the  number  of  terms  =  19)798 

Quotient  =  42  ' 
Subtract  the  first  term  =    3 

Or,  -^^^^^  —  3=^39. 

19 
2d.   A  merchant  being  indebted  to  12  creditors  ^2460,  ordered 
his  clerk  to  pay  the  first  ^40,  and:  the  rest  increasing  in  arithmeti- 
cal progression:   i  demand  the  difference  of  the  payments,  and 
the  last  payment  ? 

Ans.  g3i)=diff";  and  ^370.  last  payment. 

Problem  X. 
Given  the  last  term^  the  number  of  terms,  ai(d  the  mm  of  the  termSf 
to  find  the  first  term. 
Rule. 
Divide  twice  the  sum  by  the  number  of  terms  ;  from  the  quo- 
tient subtract  the  last  term,  and  the  remainder  will  be  the  first. t 

Examples. 
1.  If  the  last  term  be  39,  the  number  of  terms  19,  and  the  sum 
(if  the  series  399  ;  what  is  the  first  term  ? 

Sum  of  the  series  =  399 
Multiply  by       2 


Divide  by  the  number  of  terms  ==  19)798 

Quotient  =  42 
From  the  quotient  take  the  last  term  =  39 

Or,  ..^^^^^  —  39=3.  Remainder  =    3  Ans. 

19 

"=  By  Prbb.  II.  the  product  of  the.sum  of  the  series  and  the  number  of  term?, 
divided  by  2,  gives  the  sum  of  the  series  ;  whence  twice  the  sum  of  the  series 
divided  by  the  number  of  terms,  and  the  quotient  diminished  by  tlie  first  term, 
will  ^We  the  last  term. 

t  By  Prob.  II.  half  the  product  of  the  sum  of  the  extremes  and  the  number 
of  terms,  gives  the  sum  of  the  terms  ;  whence,  twice  the  sum  of  the  terms  di- 
vided by  the  number  of  terms,  and  the  quotient  diraiaished  by  the  last  term, 
must  give  ^he  first  term. 


228  ARITHMETICAL  PROGRESSION. 

2.  A  man  had  10  sons,  whose  several  ages  differed  ^like  ;  the 
eldest  was  48  years  old,  and  the  sum  of  all  their  ages  was  255  : 
What  was  the  age  of  the  youngest  ?  Ans.  3  years. 

Problem  Xf. 

The  common  difference,  number  of  terms ,  and  the  last  term  given,  to 
find  the  first  term. 

Rule. 
From  the  last  term  subtract  the  product  of  the  terms  less  1  bv 
the  common  difference,  and  the  remainder  will  be  the  first  term.* 

Examples. 
1.  If  the  common  difierence  be  2,  the  number  of  terms  19,  and 
the  last  term  30  ;  what  is  the  fir?t  ?  Last  term  =  39 

Subtract  the  number  of-terms  less  1  >  __-j-  o  —  q/> 

multiplied  by  the  common  difference    S  ~  ~ 

Remains  3  An?, 


Or,  39—19—1  x^  =  3. 
2.  A  man  travelled  6  days,  each  day  going  4  miles  farther  than 
on  the  preceding  day,  till  the  last  day's  journey  was  40  miles  ;  how 
far  did  he  ride  the  first  day  ?  '      Ans.  20  miles. 

Problem  XII. 

The  common  difference^  tJie  number  of  terms ^  and  last  term  given,  to 
find  the  sum  of  the  series. 

Rule. 
From  the  last  term  take  the  number  of  terms  minus  1,  multipii^ 
ed  by  half  the  common  difference,  and  the  remainder,  multiplied 
by  the  number  of  terms,  will  give  the  sum-t 

Examples. 
1.  If  the  common  difference  be  2,  number  of  terms  19,  and  tho 
last  term  39  ;  what  is  the  sum  of  the  series  ?     Last  term=39 

Subtract  the  number  of  terms  [ess  1  ?  __,y ,       i  =  ip 

multiplied 'by  |  the  common  difference  5  — 

Remainder  =  21 
Multiply  by  the  number  of  terms  =  19 

189 
21 

—  Answer,  399 


Or,  19x39—19—1  X  1=399 

-'  By  Prob.  I.  the  difference  of  the  extreme?  divided  by  the  number  of  terms 
less  1,  gives  the  common  difference,  whence  the  last  term  diminished  by  the 
product  of  the  common  diflerence  and  the  number  of  terms  less  1,  must  give 
the  first  term.  < 

t  By  Prob.  XI.  find  the  f.rst  term,  and  then  by  Prob.  VIII.  find  the  sum  of  the 
pi'ogression.  The  rule  is  only  a  contraction  of  this  process,  as  may  be  seen  in 
working  an  example,  and  keeping  the  several  terms  separate  in  the  operatioQ. 


ARITHMETICAL  PROGRESSION.  ^29 

2.  A  man  performed  a  journey  in  6  days,  and,  each  day,  travell- 
ed 4  miles  farther  than  on  the  preceding  day,  till  his  last  day's 
travel  was  40  miles  ;  how  far  did  he  travel  in  the  whole  ? 

Ans.  180  miles. 
Problem  XIII. 
TJie  sum  of  the  terms,  the  number  of  terms,  and  the  common  diff'er- 
ence  given,  to  find  the  first  term. 
Rule. 
Divii^  the  sum  by  the  number  of  terms  ;  from  the  quotient  talie 
half  the  product  of  the  number  of  terms,  minus  unity,  by  the  com- 
mon difference,  and  the  remainder  will  be  the  first  term.* 

Examples. 

1.  If  the  sum  of  the  series  be  399,  the  number  of  terms  19,  and 
the  common  difference  2  ;  what  is  the  first  term  ? 

Number  of  terms  =  19)399==sum- 

Quotient  =  21 
Subtract  |  the  product  of  the  number  of>  __  ■    9  —  i<^ 

terms,  less  1,  by  the  common  difference  \      -! ~ 

^     Ans.  3 

399  2x  19—1 

2.  A  man  travelled  180  miles  in  G  days  ;  he  increased  his  jour- 
ney, each  day  by  4  miles  :  how  far  did  he  travel  the  first  day  } 

Ans.  20  mites. 
Problem  XIV\ 
TJie  sum  of  the  termSy  number  of  terms,  and  the  cormnon  difference 
given,  to  find  the  last  term. 
Rule. 
Divide  the  sum  of  the  series  by  the  number  of  terms  ;  to  the 
quotient  add  half  the  product  of  the  number  of  terms  minus  unity 
by  the  common  difference,  and  the  sum  will  be  the  last  term.j 

Examples. 
1.   If  the  sum  of  the  series  be  399,  the  number  of  terms  19,  and 
the  common  difference  2  ;  what  is  the  last  term  ? 

Divide  by  the  number  of  terms  =  19)399  sum. 
Quotient  =  21 
Add  1  the    product  of  the  number  of/  __19 — 1x2 
terms,  Jess  1,  by  ihp  common  difference  J  ^       =1C 

399     2Xi>-»— T  Ans.  =39 

*  By  Pro!-).  VIII.  the  product  of  the  number  of  terms  less  !,  ftntl  of  half  the 
common  difference,  added  to  the  first  term,  and  the  sum  multijjlied  by  the  num- 
ber of  terms,  gives  the  sum  of  the  progression,  whence  divide  the  sum  of  the  se- 
ries by  the  number  of  terms,  and  diminish  the  quotient  by  the  product  of  th« 
number  of  terms  less  ]  and  half  the  common  difference,  or  by  half  tlie  product 
.  of  the  number  of  terms  less  1  and  the  common  diflerence,  and  you  have  the 
first  term. 

t  This  rule  is  obtained  from  the  rule  of  Prob.  XII.  in  a  similar  manner  as 
the  preceeding  rule  from  Prob,  VIII. 


2'SO 


ARITHMETICAL  PROGRESSION. 


2.    A  person  bought  a  farm  for  £510  to  be   paid  monthly  in 
arithmetical  progression,  and  to  be  completed  in  a  year,  each  pay 
ment  to  exceed  that  preceding  by  £5:    What  were  the  first  and 
last  payments  ? 

Ans.  £15  the  first  payment,  and  £70  the  last  payment. 

T^e  following  Table  contains  a  summary  of  the  "isihole  doctrine  of 
Arithmetical  Progression. 

Note.  The  table  contains  several  cases,  whose  rules  are  not 
given  in  the  text,  because  they  are  not  very  easily  demonstrated 
without  the  aid  of  Algebra.  Each  of  these  cases  however  is  illus- 
trated by  examples,  which  follow  the  table,  so  that  the  expresjiion 
for  the  process  in  the  table,  may  be  more  intelligible  to  the  learner. 

It  should  be  observed  that  where  two  letters  or  a  figure  with  a 
letter  or  letters,  occur  in  the  rules  in  the  table,  without  a  sign  be- 
tween them,  the  product  of  the  quantities  is  intended.  Thus,  d(> 
means  t/Xo,  and  ^ds  signities  8Xc/X5. 


CASES  OF  ARITHMETICAL  PROGRESSION, 


Case  I    Givfcfi     I  KftjViir-o 


J?(/IullOfi 


a,  /,  n, 


/ — a 


n— 1 


Prob.  I. 


a-\-lxn 


Prob.  II, 


3. 


a,l,^''d 


a,  /,  s,  <  -- 


I — a 


+  1   Prob.  III. 


l-{-aXl — a-\~d 

"  2d 


Prob.  IV 


-,-;     Prob.  V. 


l-\-ax  l~a 

^-=:z  Prob.  VI. 

2s~lJf.a 


4. 

ct,  d,  s,  <  - 

n 

2d 

I 

x/2a- <i\" -\-Us-d 

2 

ARITHMETICAL  PROGRESSION. 


2m 


Ca  !    I   oiven   |    Required 


•^.ij;  lion 


a,  dy  n, 


a,  7*,  i", 


/,  dy  5, 


I 


n—lxd+a  Prob.  VII. 


nXa+?j— IX- 


Prob.  VIII, 


2xs  -an 


n—lxn 


2s 


-a     Prob.  IX. 


d±_^2lJrd\'—^ds 
'  2 


^i-\-d-\-y/n-\-d\^—Us 


2d 


8. 


/,      n,       Sy 


zs 


■I     Prob.  X. 


2  X  nl — s 
n—lxn 


d. 


I,  7i,  d. 


l—n — IXc^ 


Prob.  XI. 


nxl-n~lXd     p^^^   XII. 

2 


10. 


dyUyS, 


dXn- 


Prob.  XIII. 


_  J Prob.  XIV. 

n    '  2 


a  =  farsi  lerrn,  or  least  term. 


\/  =  la?t  term. 
Here  <«  =  number  of  te 
a  d  =  common  diffe 


rms. 
irence. 
s  =  sum  of  all  the  terms. 


232  AkITHMETICAL  PROGRESSlOxV. 

Examples  in  Arithmetical  Progression. 

1.  Given  the  first  term  3,  the  common  difference  2,  and  the 
sum  of  the  series  399,  to  find  the  number  of  terms  and  the  last 
term. 

Bv  the  4th  Case  in  the  table  we  have, 


2x2 


And,  V  3x2-21   -|-399x2x8--2^3^^  ^^^  j^^^  .^^^^ 
2 

2.  Given  the  first  term  3,  the  number  of  terms  19,  and  the  sum, 
c^the  terms  399,  to  find  the  common  difference. 

?Dy  the  first  rule  of  Case  6th,  we  have, 

2x399—3x1^ 

_______ =2,  the  common  difference. 

19—1x19 

3.  Given  the  common  difference  2,  the  iast  term  39,  and  the 
sutii  of  the  series  399,  to  find  the  firat  term  and  the  number  of 
terms. 

By  Case  7,  we  have, 

-     24-^/39x2+21^— 399x2x8  _c,    ,,  ytv 

^      ^    L-J =3,  the  common  difference;. 


And,  39x2+-2±V39x2+2r-399x2x3^,,    ^^^    ^^^^^^  ^^, 

2X2 
terms. 

4.  A  merchant  owed  to  several  persons  ^1080;  to  the  greatest 
creditor  he  paid  ^142,  to  the  greatest  but  one  ^132,  and  so  on,  in 
Arithmetical  Progression  ;  What  was  the  number  of  creditors,  and 
what  did  the  least  creditor  receive  ? 

Ans.  The  number  of  creditors  was  15,  and  the  least  creditor 
received  ^2. 

5.  Given  the  last  term  39,  the  number  of  terms  19,  and  the  sum 
of  the  series  399,  to  find  the  common  difference. 

By  the  2d  rule  in  Case  8,  we  have, 

2X19X39—399 

" — -  =2,  the  common  difference. 

19—1X19 
C.  Sixteen  persons  gave  in  charity  to  a  poor'man  in  such  a  man- 
ner as  to  form  an  arithmetical  series;  the  last  gave  65  cents,  and 
I  he  whole  sum  was  §5  60c. ;  What  did  each  give  less  than  the  oth- 
er, from  the  last  down  to  the  first. 

Ans.  4  cents. 


QEOMETRICAL  PROPORTION'. 


GEOMETRICAL  PROPORTION. 

Theorem  L 

IF  four  quantities,  2.  6  4  12,  be  in  Geometrical  Proportion,  the 
product  of  the  two  means,  6X4  will  be  equal  to  that  of  the  two  ex- 
tremes, 2X12,  whether  they  are  continued,  or  discontinued,*  arul, 
if  three  quantities,  2.  4.  8,  the  square  of  the  mean  is  equal  to  the 
product  of  the  two  extremes. 

Theorem  2. 

If  four  quantities,  2.  6.  4.  12,  are  such,  that  the  product  of  two 
of  them,  2x12  i«  equal  to  the  product  of  the  other  two,  6x4,  then 
are  those  quantities  proporlional.t 

» 

'^^  It  wai  statsd  uriler  Proportion  iq  General,  that  the  geometrical  ratio  of  two 
quantities  is  expressed  by  the  quotient,  arising  from  dividing  the  antecedent  by 
the  consequent ;  thus,  the  geometrical  ratio  of  G  to  2  is  3,^=  p  ^  and  of  2  to  6,  is 
f  or  -g-^  and  of  3  to  8  is  | :  and  that  in  a  proportion  there  must  be  two,  or  more, 
couplets  which  have  equal  ratios.  Hence,  four  numbei-s  will  be  geometrical  pro- 
portionals,  when  the  ratios,  obtained  in  this  manner,  arc  equal.  Thus  2, 4, 8,  16, 
are  geometrically  proportional,  because  2=:_8„:  each  to  i  ;  and  thus  also,  9, 
o,  12,  4,  because  f=V^r^each  to  3.  From  these  principles,  it  is  easy  to 
prove  in  a  given  example,  the  theorems  in  Geometrical  Proportion.  The  fac- 
tors should  be  kept  separate  by  the  sign  of  multiplication. 

Let  2, 4, 3, 6,  be  the  geometrical  proportionals  ;  then  1 =3^  Multiply  both  frac- 
tions by  the  product  of  the  second  and  fourth  terms,  and  the  fractions  will  obvi- 
ously still  be  equal,  and  we  have — ==,■ '■ — .      Then  cancel  the  equal 

4  6 

terms  in  the  fractions,  and  2  X  6=3  X  4,  that  is.,  the  product  of  the  extremes,  2  X  C, 
fs  equal  to  the  product  of  the  means,  4X3.  The  same  may  be  shown  ia  any 
other  case,  and,  hence  the  general  rule  be  inferred. 

Again  ;  Three  numbers  are  geometrical  proportionals,  when  the  ratio  of  the 
first  and  second  terms  is  equal  to  the  ratio  of  the  second  and  third.     Thus, 
2,  4,  8,  are  three  geometrical  proportionals,  for  2:r:4=each  to  ^.     Proceed  as 
2x4x8      4X4X8  * 

before,  and  we  have-^^^ — — ,  and  2X8=4X4,  or  4^,  that  is,  the 

4  o 

product  of  the  extremes,  2X8,  is  equal  to  the  square  of  the  mean,  4X4  or  4^- 

tLet  the  four  quai^tities  be  2,  6,  4,  and  12,  so  that  2x12=6x4.  Di-.ido 
these  equal  products  by  the  quantity  6X12,  and  the  quotients  will  obviousK- 

2x12      6x4 
be  equal,  or ^z=:- — — .     Cancel  the  equal  terms  in  these  two  fractions,  and 

we  have  f =p\  ;  whenes  2  :  6  :  :  4  :  12,  by  the  defmition  of  geometrical  pro- 
portionals. 

In  the  same  way  it  may  be  shown,  that  if  2x8=nE4x4  or  4^,  then  2:4::  4  :  8, 
and  3,  4,  and  8,ar3  thre^  goometrioul  proportionaT'i. 

F   f 


2 

6 

::      4  : 

12 

6 

2 

::    12  : 

4 

o 

4 

:;     6  : 

12 

2 

8 

::     4  : 

16 

2  : 

4 

•:     4  : 

8 

8 

:  4 

::    16   : 

8 

234  GEOMETRICAL  PROPORTION. 

Theorem  3. 
If  four  quanllties,  2.  6.  4.  12,  are  proportional,  the  product  of 
the  means,  divided  by  either  extreme,  will  give  the  other  extreme.*^ 

TiTEOREM  4. 

The  products  of  the  corresponding  terms  of  two  Geometrical 
Proportions  are  also  proportional. 

That  is,  if  2  :  6  i  :  4  :  12,  and  2  :  4  :  :  5  :  10,  thea  will  2x2  : 
6X4  :  :  4x5  :  12xl0t 

Theorem  5. 
If  four  quantities,  2,  6,  4,  12,  aie  directly  proportional. 
(  1.   Directly, 

2.  Inversely, 

3.  Alternately, 

4.  (  ompoundedlyj 
Then,  {  5.   Dividedly, 

6.  Mixtly, 

7.  By  Muitiplltalion,  2X5  :  6X3  ::  4  :  12 
2      6 

8.  By  Division,  -  :  -    ::    4  :   12 
I  3      5 

Because  the  product  of  the  means,  in  each  case,  is  eqnal  to  that 
of  the  extreujes,  and  therefore  the  quantities  are  proportional  by 
Theorem  1. 

Theorem  6. 

If  three  numbers,  2,  4,  8,  be  in  continued  proportiou,  the  square 
of  the  first  will  be  to  that  of  the  second,  as  the  first  number  to  the 
third  ;  that  is,  2X2  :  4X4  ::  2  :  84 

*  Let  the  four  proportionals  be  2,  4,  5,  and  10  ;  then  2X10=4X5,  by  Theo- 

2X10     4X5  4X5 

rem  1.     Divide  both  expressions  by  2,  and  — - — =— -- ;  ol-  10=—^— ;  ^  or,  di- 

2x10     4X5  4X5 

vide,  as  before,  by  10,  and  ■   r=----,  or  2= — ^--,  that  is,  the  product  of  the 

means  divided  by  one  extreme,  gives  the  other  extreme.  Hence,  if  the  two 
means  and  one  extreme  be  ^iven,  the  other  extreme,  or  geometrical  proportion- 
al may  be  found. 

t  Let  there  be  given,  2  :  6  ::  4  :  12,  vi'hence  ^^r-^-  by  Theorem  1  ;  and  also, 
3  :  5  ::  6  :  10,  whence  |  =  ro-     Multiply  the  corresponding  parts  of  these  equal 

2X3             6x4 
fractions  tojrether,  and  we  have  ~  —  and  - — ■ and  these  products  are  obvi' 

2X.i       6X4     ^>^'^  l^X^^^ 

©usly  equal,  or =. — — —     Hence  by  the  definition  of  geometrical  propor- 

^    ^  6X5     10X12  ^  *= 

tionals,  2X3:6X5::  6x4  :  lOx  12,  and  is  the  theorem. 

•  .Hence,  if  four  quantities  are  proportional,  their  squares,  cubes,  &c.  will  likewise 
beproportional.  Thus,  let  the  terms  be  2  :  6  ::  4  :  12,  then  2x2  :  6  X6  ::  4x4 
:  12X12,  or  2^  :6^   ::  4^  :  12",  and  hence  also,  2^  :6^   ::  4^  :  12^  and  2*  ; 

6^   ::  4^  ::12'. 

:j:  For  since  2  :  4  ::  4  :  'fi.  thence  will  2X8=:4x4,  by  Theorem  1  ;  and 
therefore  2  X  2  X  C=2  X  4  X  4,  by  equal  m ultiplication ;  consequently,  2^2:4x4 
::  2  :  8,  by  Theorem  2. 

In  like  manner  it  may  be  proved  that,  of  four  quantities  continually  propor- 
tional the  mtlfc  of  the  iu\X  k  to  llial  of  the  second,  as  the  first  quantity  to  the 
foyjrth. 


GEOMETRICAL  PROGRESSION.  2^ 

TliEpREM    7. 

In  any  continued  Geometrical  Proportion,  | ,  3,  9,  27,  81 ,  &c.  the 
product  ofthe  two  extremes,  and  that  of  every  clher  two  term^ 
equally  distant  from  them  are  equal.* 

Theorem  8. 

The  sum  of  any  number  of  quantities,  in  continued  Geometrical 
Proportion,  is  equal  to  the  difference  ofthe  product  ofthe  second^ 
and  last  terms,  and  the  square  of  the  tirst,  divided  by  the  differ- 
ence ofthe  first  and  second  terms.! 


GEOMETRICAL  PROGRESSION. 

A  GEOMETRICAL  Progression  is,  when  a  rank  or  scries  of 
numbers  increases,  or  decreases,  by  the  continual  multiplication, 
or  division,  of  some  equal  number,  which  i$  called  the  ratio. 

^  For,  the  ratio  of  the  .first  term  to  the  second  being  the  same  as  that  of 
f.he  last  but  one  to  the  last,  these  four  terms  are  in  proportion ;  and  there- 
fore by  Theorem  1,  the  product  ofthe  extremes  is  equal  to  that  of  their  two  ad- 
jacent terms  ;  and  after  the  same  manner,  it  will  appear  that  the  product  of  the 
third  and  last  but  two  is  equal  to  that  of  their  two  adjacent  teims,  the  sec- 
ond and  last  but  ontj,  and  so  of  the  rest ;  whence  the  truth  of  the  proposi- 
tion is  manifest. 

t  Take  any  series  of  continued  geometrical  proportionals,  as  2,  6,  18,  51, 
162,  486,  and  its  sum  is,  2+6-f  lb-|-54H-162H-486,=728,  by  addition.  Multi- 
ply the  whole,  by  that  number  by  which  any  term  of  the  series  is  multiplied  or 
4ivided  to  form  the  succeeding  term,  which  is  in  this  example,  3,  and  you  have, 
..6+184-54+162+486+1458=2184.  Subtract  the  first  series, 
2+6+18+54+162+486  -  -  -  ::r^728.  As  all  the  terms  in  the  up- 
per series  are  cancelled  by  those  in  th§  lower,  except  the  last  in  the  former 
and  the  first  in  the  latter,  those  two  terms  become — 2+1458,  or,  1458— 2=r 
2184 — 728.  Now,  the  iipper  series  exceeds  the  lower  three  times,  and,  of  course, 
from  thrice  the  series  there  has  been  taken  once  the  series,  and  the  remain- 
der must  be  iicice^  or  3 — 1  times,  the  series  and  equal  to  twice  or  3r-rl.tiiHe?, 
the  sum  of  the  series,  that  is,  -~^ — r-  =  728,  the    sum  of  the   series.     As 

^^53__2  3X486 2 

-^ is   also   — pr ^,  multiply  botli  parts  of  the  fraction  by  2,  which. 

•^r\    ,.     ..       1  ^^\         1         2X3X486—2X2        6x486—4        ... 
Will  not  alter  its  value,  and  you  have : — ; — or  ,  winch 

2x3—1  ^~^ 

must  also  be  equal  to  tlie  sum  (jf  the  series,  and  is  the  rule.     For  6  is  the 

second  temx  of  the  given  series ;  486,  the  greatest  term ;  4,  the  square  of  tlio 

second  term ;  £^nd  the  divisor,  6 — 2,  is   the  difference  of  the  first  and  scconcT 

terms. 

3X486 2 

In  this  deniQustration  it  is  shown  that  '—-. j; — =rtl)esum  of  the  series ;  that 

js,  that  the  greatest  term  multiplied  by  the  number  by  which  the  series  in- 
creases, and  the  product  diminished  by  the  least  term,  and  this  divided  by  e 
number  one  less  than  that  by  which  the  series  increases,  tlsa  quotient  is  tJie 
sum  of  the  series.     Let  the  series  be  1,  4,  16,  64,  256,  1024,  whose  multi? 

4X10'?  4 1  " 

plicr  i3  4.     Then  — — ^—1365,  the  rum  of  the  series. 


236  GEOMETRICAL  PROGRESSION. 

Problem  I. 

Given  one  of  the  extremes^  the  ratio^  and  the  number  of  the  terms  of 
a  geometrical  series^  to  find  the  ether  extreme. 

Rule. 
Multiply  or  divide,  (as  the  case  may  require)  the  given  extreme 
by  such  power  of  the  ratio,  whose  exponent*  is  equal  to  the  num- 
ber of  terms  less  1,  and  the  product  or  quotient,  will  be  the  other 
extreme.! 

*  As  the  last  term  or  uny  term  near  the  last,  is  very  tedious  to  be  found,  by 
continual  multiplication,  it  will  often  be  necessary  in  order  to  ascertain  them,  to 
have  a  series  of  numbers  in  Arithmetical  Proportion,  called  indices^  or  exponents^ 
beo^innin^  with  a  cypher,  or  a  unit  whose  common  difference  is  one. 

When  the  first  term  of  the  series  a-nd  the  ratio  are  equals  the  indices  must 
begin  with  a  unit,  and,  in  this  case,  the  product  of  any  two  terms  is  equal  to 
that  term  signified  by  the  sum  of  their  indices.  This  is  obvious  on  inspection  of 
this  example. 

rpi  <  1.  2.  3.     4.     5.     6,  &;c.  indices,  or  arithmetical  series. 

'    (  2.  4.  8.  1 6.  32.  64,  &;c.  geometrical  series  (leading  terms.) 

Now,  64-6=:12=the  index  of  the  twelfth  term,  and 
64X64=4096— the  twelfth  term. 

But,  when  theirs/  term  of  the  series  and  the  ratio  are  different^  the  indices 
must  begin  with  a  cypher,  and  the  sum  of  the  indices.,  made  choice  of,  must  be 
one  less  than  the  number  of  terr^u^  given  in  the  question  ;  because  1  in  the  indices 
stands  over  the  seeond  ttrm^  and  2  in  the  indices.,  over  the  third  term,  &;c.  And, 
in  this  case,  the  product  of  anV  two  terms  divided  by  the ^r5/,  is  equal- to  that 
term  beyond  the  first,  signified  by  the  sum  of  their  indices^  as  is  obvious  from 
this  example. 

rp,  <  0.  1.  2.    3.     4.      5.       6,  &c.  indices. 

'    ^  1.  3.  9.  27.  81.  243.  729,  &c.  geometrical  series. 

Here,  6  +  5  =  11  the  index  of  the  12th  term. 

729X243=177147  the  12th  term,  because  the  first  term  of  the  series 
and  the  ratio  are  different,  by  which  means  a  cypher  stands  over  the  first  term. 

Thus,  by  the  help  of  these  indices,  and  a  few  of  the  first  terms  in  any  geomet-. 
yical  series,  any  term,  whose  distance  from  the  first  term  is  assigned,  though  it 
were  ever  so  remote,  may  be  obtained  without  producing  all  the  terms. 
• 

t  The  rule  is  evident  from  the  manner  in  which  a  geometrical  progi-ession  i^ 
formed.  Let  2,  4,  8, 1 6, 32,  Sic.  be  the  series,  whose  ratio  is  2.  The  second  term 
3ii  formed  by  multiplying  the  first  term  by  the  ratio ;  the  third  term,  by  multiply- 
ing the  second  by  the  ratio,  and  so  on.  The  series  may  therefore  be  written  thus, 
2,^2X2,  2X2X2,  2X2X2X2,  2X2X2X2X2,  &c.  or  thus  2,  2x2i,  2X:^^, 
2X2^,2X24,  and  so  on.  Any  term  after  the  first  is  evidently  that  power  of  the 
ratio  whose  index  is  one  less  than  the  number  of  the  term  multiplied  by  the  first 
lertn.  Thus,  the  3d  term  is  2X23  ;  the  4th  term  is  2X2^,  and  the  8th  term 
would  be  2X2 "7,  and  so  on.  In  an  ascending  series, iherefore^  multiply  the  first 
term  by  that  power  of  the  ratio  whose  index  is  one  less  than  the  number  of  the 
term  souglit,  and  the  product  is  the  term  bought. 

In  a  descending  series,  as  243,  81, 27, 9, 3, 1,  whose  ratio  is  3,  and  which  is  also 

1X35 
1  X3»,  1X31,1  X33,  1 X32, 1 X31, 1,  the  last  term,  1,  =^-— — .   Hence  the  rule 

li  evident,  whatever  be  the  fir?i  term  or  the  ratio. 

JSTote  I.  If  the  ratio  of  any  geometrical  series  be  2,  the  difference  of  the  great- 
est and  least  terms  is  equal  to  the  sum  of  all  the  terms  exfiept  tlie  greatest ;  if  the 
ratio  be  3,  the  difference  is  double  the  sum  of  all  the  terms  except  the  greatest; 
if  the  ratio  be  4,  tlie  difference  is  triple  the  sum,  itc.  Let  the  series  be  1,2,  4,  8, 
10,  32,  whose  ratio  is  2,  and  whose  sum  is  63,  and  the  sum  of  whose  fii"st  five 
terms  is  31.     Now  32—1,  the  differeiice  of  the  extremes,  is  e^'wa/ to  the  suim  > 


^ 


GEOMETRICAL  PROGRESSION.  237 

Examples. 

1.  If  the  first  term  be  4,  the  ratio  4,  and  the  rrnmber  of  terms  9  ; 
Wliat  is  the  last  term  ? 
1.     2.     3.     44-     4=     8 

4.  16.  64.  256.x256=65636=power  of  the  ratio,  whose  exponent 
•3  less  by  1,  than  the  number  of  terms. 

65536=8th  power  of  the  ratio 
Multiply  by  4=firsl  term. 


262144=last  term. 
Or,  4x48=262144=the  Answer. 
2.   If  the  last  term  be  262144,  the  ratio  4,  and  the  number  of 
terms  9,  what  is  the  first  term  ?  4=lhe  first  term. 

the  first  five  terms.     For,  by  the  principle  proved  under  Theorem  8,  of  G^omet- 

i>Xl6=:l         32 I 

rical  Proportion,  the  sum  of  the  first  five  terms  is         ,  or r=32 — h 

the  difference  of  the  ^eatest  and  least  terms. 

Let  the  series  be,  3, 9,  27,  81,  243,  whose  ratio  is  3,  then  243--3=twice  th« 

^JX81 — 3 
sum  of  all  the  terms  except  the  greatest.     Now  by  the  same  principle  — ^ 

243 3    243- 3 

=sum  of  all  the  terms  except  the  last,  =.--- — —=: — ,  that  is,  the  sum  of  all 

3 — 1  2 

,  the  terms  except  the  greatest  is  half  the  difference  of  the  greatest  and  least  terms, 

or  this  difference  is  twice  the  sum  of  all  the  terms  except  the  greatest. 

Let  the  series  be  1,4, 16, 64, 256,  whose  ratio  is  4,  as  before =  the 


05^ 1      256 1 

sum  of  all  the  terms  except  the  greatest,  =^- 1=--— 1  or  tbe  difference 

of  the  greatest  and  least  terms  is  equal  to  thrice  the  sura  of  the  series  except  the 
last  term. 

JVo/e  2.  In  any  geometrical  progression  decreasing;  to  infinity,  or  beyond  any 
Hssignable  limit,  the  square  of  the  Jirst  terrn,  divided  by  the  difference  between  ih& 
first  and  second  terms^will  be  the  sum  of  the  series. 

Let  the  series  be  i,  1,  1,  -'-,  &:c.  to  infinity.     By  Theorem  8  of  Geometrical 

Proportion,  the  sum  of  any  number  of  terms,  an  the  first  four  terms,  is  ^—  ^  ^  yy 
-i-i,  or  frohn  the  equare  of  the  firat  term  the  pi-oduct  of  the  seciond  and  fourth 
terras  is  to  be  taken  and  the  remainder  divided  by  the  difference  of  the  first  and 
second  terms.  But  if  the  series  be  infinite,  the  last  term  is  infinitely  small  and 
must  be  considered  0 ;  and  then  the  product  to  be  taken  from  the  square  of  th<» 
first  term  is  0..  For  when  _L.  i?  supposed  infinitely  small  or  0,  then  {  X  J-  be- 
oomes  ixO=::0;  and  the  expression  becomes  l-^^i, -or  ^Xi=l, the  sumof  the 
above  series  continued  without  end.     Hence  the  rule  is  manifest. 

The  above  expression  ^ — |XJ~^isals6^ — 'I'^tV"^-^'  tbat  \s,  multiph^ 
the  last  term  by  the  raiio^i  and  divide  the  difference  between  this  product  and  the 
first  term  by  the  difference  between  1  and  the  ratio^  and  the  quotient  Avill  be 
the  sum  of  the  series.     Let  the  series  be  1 ,   i^  j.   jl^   whose  sum  is,  by  addition.; 

'^f  Tiien  by  the  rale,  THi X  J^-f-l  —1=  1  _1  xy-^~|,=|f  X|=:^|-, 
as  before.  If  the  series  were  continued  infinitely,  then  as  before  the  product  to 
be 'subtracted  from  thefirst  term  would  be  (}•,  and  ]_f.2=;lx^=l"  ^'^ald  be 
the  ^um.  That  is,  if  the  series  descend  to  infinity,  divide  the  frsf  fna  by  the 
difference  between,  unify  and  (he  ratio,  and  the  quotif  ut  will  be  tlio  c  .vu  of  th« 


2Sii  GEOMETRICAL  PKOGRESSION. 

Jigain^  given  the  first  terrn,  and  the  ratiOf  to  find  any  other  tena 
y  assigned. 

Rule  I  * 
When  the  indices  begin  with  a  unit. 

1.  Write  down  a  few  of  the  leading  terms  of  the  series,  and 
place  their  indices  over  them. 

2.  Add  together  such  indices,  whose  suai  shall  make  up  the  en- 
tire index  to  the  term  required. 

3.  Multiply  the  terms  of  the  geometrical  series,  belonging  to 
those  indices,  together,  and  the  product  will  be  the  term  sought. 

Examples. 

1.  If  the  first  term  be  2,  apd  the  ratio  2,  what  is  the  13th  term  ? 

1.  2.  3.     4.     64-6x3=     13 

2.  4.  8.   \Q.  32x32x8=8192  Ans.  Or^  2x2^2=^8192. 

2.  A  merchant  wanting  to  purchase  a  cargo  of  horses  for  the 
West-Indies,  a  jockey  told  him  he  would  take  all  the  trouble  and 
expense  upon  himself,  of  collecting  and  purchasing  30  horses  for 
the  voyage,  if  he  would  give  him  what  the  last  horse  would  come 
to  by  doubling  the  whole  number  by  a  halfpenny,  that  is,  two  far- 
things for  the  first,  four  for  the  second,  eight  for  the  third,  &c.  to 
which,  the  merchant,  thinking  he  had  made  a  very  good  bargain, 
readily  agreed  :  Pray  what  did  the  last  horse  come  to,  and  what 
did  the  horses,  one  with  another,  cost  the  merchant? 

1.  2.  3.     4.     5.     6-f-  6=12th.  12-f    12+  6=last  term. 

2,  4.  8.   16.  32.  64x64=4096,  and  4096x4096x64  = 
1073741824qrs.=  £11 18481   Is.  4d.  and  their  average  price  war^ 
£37282   14s.  O^d.  a  piece. 

RuleII.* 

When  the  indices  begin  -with  a  CTj-pher. 
1.   Write  down  a  few  of  the  leading  terms  of  the  series,  as  be 
fore  and  place  their  indices  over  them. 

iaihiite  series.     Thus,  the  sum  of  — --, , ,  — --- ,&c.  to  hiTiuIiv 

1-05 '    1-06:^'    1-U53'    1-06  i 


1111 

And  the  sum  of  four  tcriiis  of  the  series, ,  ,  ,  — — 

1-05      1-053      1-053'   1-05' 


1-05       1-055  •         iW^  ^^^    ^^^'"1-05"    1-055  ■      1-05    ^   ^^     ^1-05     1-055 

1-05        ,  •  u  •    .  i  1  ,  .       .  ^  1  1  1 

X '-I  which  IS  1 X ,  which  is  1 —  ■ X  — ,= 

1-05—1  l-Uo4  '^  iU5— 1'  ■  1-054  ^-  -05'     -05       05 


^  Thcsf^  rules  sire  only  particular  cases  of  the  preceding  gjeneral  rul?,  and  th<i 
jjc&sonof  tbcm  is  obvious  from  the  dcnxonstration  of  thr.t  rule. 


GEOMETRICAL  PROGRESSION.  239 

i^.  Add  together  the  most  convenient  indices  to  make  an  indes, 
less  by  1,  than  the  number  expressing  the  place  of  the  term  sought 

3.  Multiply  the  terms  of  the  geonrietrical  series  together,  be- 
longing to  those  indices,  and  make  the  })roduct  a  dividend. 

4.  Raise  the  first  term  to  a  power,  whose  index  is  one  less  than 
the  number  of  terms  multiplied,  and  make  the  result  a  divisor,  by 
which  divide  the  dividend,  and  the  quotient  will  be  that  term  t^- 
yond  the  first,  signified  by  the  sum  of  those  indices,  or  the  term 
sought. 

6.  If  the  first  term  be  5,  and  the  ratio  3  ;  what  is  the  7th  term  ? 
0.  1.  2.  3.-f  2-f-  1=  6— ind. to 6th  term  beyond  the  1st  or 7th 
5.  15.45.  135.x45Xl5=:91125=dividend. 

The  number  of  terms,  multiplied,  is  3  (viz.  135X45x15,)  and 
3 — 1=2  is  the  power  to  which  the  term  5  is  to  be  raised  ;  but  the 
2d.  power  of  5  is  5X5=25,  and  therefore  91l25-r-25i=3645  the 
7th  term  required. 

Problem  II. 
Given  the  first  term,  the  ratio,  and  nnmhtr  of  terms  to  find  the  sum 

of  the  series. 

Rule. 

Raise  the  r^tio  to  a  power,  whose  index  shall  be  equal  to  the 

number  of  terms,  from  which  subtract  1  ;  divide  the  remainder  by 

the  ratio,  less  1,  and  the  quotient,  multiplied  by  the  first  term,  will 

give  the  sum  of  the  series.* 

Examples. 
1.  If  the  first  term  be  5,  the  ratio  3,  and  the  number  of  tenmi 
7  ;  what  is  the  sum  of  the  series  ? 

Ratio=3X3x3x3X3X3x3=2187=i:7th  power  of  the  rat'c. 
Subtract  1 

t)ivide  by  the  ratio  less  1=3:3—1=2)2185 

Q,uotient=Ti093 
Multiply  by  the  first  term  =       5 

Sura  of  the  senes=5465 

37—1 
Or,    g_^-X5=5465  Aos. 

*  This  rule  is  a  contraction  of  the  following  process.  Find  by  Problem  f. 
live  last  term  of  the  progression,  and  then  find  by  Theorem  8,  of  Geometri- 
cal Proportion,  the  sum  of  the  series.  Thus  in  Ex.  1.  where  5  is  the  first 
term  and  3  the  ratio,  and  the  sum  for  7  terms  is  required.     By  Prob.  I.  the 

5X3X5X3  6 •'»  2 

7tli   term  is  5X36.    Then  by  Theorem  8,  '- — ^ — '• • — ^  =  the  sum  of  the 

5X3—5 
terms.     The  expression  5X3X5X3^  may  be  written  5X5X3x36=52x37, 
for  3x36  raises  36  to  the  next  higher  power,  or  makes  it  3?.     Also  52x3" 
— 52  is  37 — 1X52,  and  5X3 — 5  is  also  3 — 1X5,  as  is   seen  by  multiplying 

the  terms.     Therefore '■ becomes  — :r-zz — '■ —  ^r  — rr-r " 

^  X  ^— ^  3—1 X5 3—1  X5 

37—1X5        tii—i 
and,  cancellmg  the  equal  terms  in  this  fraction,  it  becomes  — - — ;; —  or  -- — - 

3 — 1  3 — i 

X5,  and   is   the   rule.     As   the  sitme  may  be  sliown  in  any  ct^**r  eramtj^e 
the  gfeneral  rule  is  obvious. 


240  GEOMETRICAL  PROGRESSION. 

2.  A  shop  keeper  sold  13  yards  of  cloth  on  the  following  terrc?^ 
viz.  2d.  for  the  first  yard,4d.  for  the  second,  8d.  for  the  third,  &,c. 
{  demand  the  price  of  the  cloth  ? 

Ol  3 J 

■^^^3j-X2=16382d.=  £68  5s.  2d.  Ans. 

.J.  A  gentleman,  whose  daughter  was  married  on  a  new  year's 
day,  gave  her  a  guinea,  promising  to  triple  it  on  the  first  day  of 
each  month  in  the  year  ;  pray  what  did  her  portion  amount  to? 

Ans.  265720  guineas. 

4.  What  debt  can  be  discharged  in  a  year,  by  paying  1  cent  the 
tirst  month,  10c.  the  second,  and  so  on,  each  month  in  a  tenfidd  pro- 
portion ?  Ans.     111111111111c  =^1111111111    lie. 

5.  A  man  threshed  wheat  9  days  for  a  farmer,  aad  arrreed  to  re- 
ceive but  eight  wheat  corns  for  the  lirst  day's  work,  64  for  the  se- 
cond, and  so  on,  in  an  eightfold  proportion  ;  I  demand  what  his  9 
day's  labour  amounted  to,  rating  the  wheat  at  5s.  per  bushel  ?* 

Ans.     153391688  corns.     Amount=£78  Os.  5id. 

6.  An  ignorant  fop  wanting  to  purchase  an  elegant  house,  a  fa- 
cetious gentleman  told  him  he  had  one  wliich  he  would  sell  him  on 
these  moderate  term?,  viz.  that  he  should  give  him  a  cent  for  the 
first  ddor,  2  cents  for  the  second,  4  cents  for  the  third,  and  so  on, 
doubling  at  every  door,  which  were  36  in  all:  It  is  a  bargain, 
cried  the  simpleton,  and  here  is  a  guinea  to  bind  it:  Pray  whfit 
did  the  house  cost  him  ? 

036 2 

Xl=68719476735c.=g6871947G7  35c.     Ans. 

2—1 

7.  A  young  fellow,  well  skilled  in  numbers,  agreed  with  a  rich 
farmer  to  serve  him  10  years,  without  any  other  reward,  but  the 
produce  of  one  wheat  corn  for  the  first  year,  atid  that  produce  to 
be  sowed  from  year  to  year,  till  the  end  of  the  time,  allowing  the 
increase  but  in  a  tenfold  proportion  ;  what  is  the  sum  of  the  whole 
produce,  and  what  will  it  amount  to  at  ^1  25c.*  per  bushel  ? 

Amounl=$22605  61. c.  3m.-f 

8.  Suppose  one  farthing  had  been  put  out  at  6  per  cent,  per  an- 
num, Compound  Interest,!  at  the  commencement  of  the  Christian 
era  ;  what  would  it  have  amounted  to  in  1784  years;  and  suppose 
the  amount  tp  be  in  standard  gold,  allowing  a  cubic  inch  to  be 
worth  531.  2s.  8d.  how  large  would  the  mass  have  been  ? 

Ans. Xl=£l486716346568748209435714551509890767065361  11  3i} 

2—1 

^=27980859722121230415979571232933594210766  cubick  inches  of  gold. 

As  355  :  113  ::  3G0xtiy5  :  7964  earth^s  diameter.      360 X69\3X 7964 X  1327-33 

=264482820122  cubick  miles  in  the  globe, 

: -6727333730885474 1368832000  cubick  inches  in  the  globe.     Then, 

27980359722121230415979571232933594210766 

*  JVute.    7380  wheat  or  barley  corns  are  supposed  to  make  a  pint. 

1-  Any  siwn  at  6  per  cent,  per  annum,  compound  interest,  "will  double  in  eleven 
years  and  three  hundred  and  t^vcnty  five  day?,  or  1  l-iiS^  year?i,  or  )  1*89  is  near 
enough ;  then,  ii  you  divide  1784  by  11-89,  it  will  give  t'l?  n-imber  of  terms  m 
this  raseonjial  to  150;  the  ratio  wiil  be  2,  and  the  first  term  1. 


<iEOMETRICAL  PROGRESSfON.  241 

'  ^6727533730885474 1368832000=?41593089984(>288-8,  which,  however  incredi- 
ble it  may  appear  to  some,  is  more  than  four  hundred  and  fifteen  millions  of  miU- 
ions,  nine  hundred  and  thirty  thousand,  eight  hundred  and  ninety-nine  millions, 
eight  hundred  and  forty  thousand,  two  hundred  and  eighty-eight  times  larger 
than  the  globe  we  inhabit.* 

For  the  solution  of  the  four  following  questions,  see  last  part  of 
note  under  Problem  I. 

9.  A  frigate  pursues  a  ship  at  8  leagues  distance,  and  sails  twice 
as  fast  as  the  ship  ;  how  far  noust  the  frigate  sail,  before  she  comes 
up  with  her?  

First,  8.  4.  2.  I.  f  i.  &c.  8x8=64,  and  64^8—- 4=16  leagues,  Ans. 

10.  Suppose  a  ball  to  be  put  in  motion  by  a  force  which  impels 
it  10  rods  the  first  minute,  8  the  second,  and  so  on  decreasing  by  a 
ratio  of  1-25  each  minute  to  infinity;  what  space  would  it  move 
through  ?  Ans.  60  rods. 

11.  Required  the  value  of -999,  to  infinity,  or  -91? 

The  first  9  or  •9,=fV,  the  second,  or  -09=^1^  ;  therefore, 

•9x9~-9-— 09=1  Ans. 

12.  Required  the  sum  of  |,  i,  i,  &c.  to  iefinity  ?  Ans.  1. 

13.  What  is  the  sum  of  i,  |,  J^*  ^^-  ^o  infinity  ?  Ans.  -i. 
H.  What  is  the  sum  of  i,  yV'  A'  ^^-  ^^  infinity  ?  Ans.  ^-. 
16.  What  is  the  sum  of -1,  -01,  -001,  &c.  to  infinity  ?    Ans.  i. 

16.  What  is  the  sum  of  1,  ^,  ^\,  ^^^  ^^-  ^^  infinity  ?    Ans.  l-J. 

17.  What  is  the  sum  off,  |-,  ^,  /y,  for  7  terms  ?     Ans.  43.12. 

1     "   1  1 

18.  What  is  the  sum  of  ^— ,  p^.j,f^  ^c.  to  mfinity  ? 

Ans.  16|. 

Problem  III. 

The  fust  term,  the  last  term  (or  the  extremes)  and  the  ratio  given^  tb 
find  the  sum  of  the  series.^ 

Rule  I. 
Divide  the  difference  of  tlie  extremes  by  the  ratio  less  by  1  ; 
add  the  greater  extreme  to  the  quotient,  and  the  result  will  be  the 
■sura  of  all  the  terms. 

Rule   11. 
Or,  Multiply  the  greatest  term  by  the  ratio,  from  the  pfoduct 
jjubtract  the  lea^^t  term  ;  then  divide  the  remainder  by  the  ratio, 
less  by  1,  and  the  quotient  will  be  the  sum  of  all  the  terms. 

**  To  fiiil  the  solid  content  of.it  globe.-  See  Art.  34th.. of  Mensuration  of 
Solids.  Note,  that  -523598  is  two  thirds  of  -785398  the  area  of  a  circle,  whose 
diameter  is  1. 

t  It  will  be  ^een,  when  we  come  to  rirrulating  decimals,  tli^t  -9  Ls  the  manner 
of  expressing  -999,  fee,  to  infinity. 

%  Rule  1.  In  Note  1,  under  Prob.  I.  it  is  shown  that  the  difference  of  the 
reatest  and  least  terms  divided  by  tiie  ratio  less  1,  gives  the  sum  of  the  series  ex- 

pt  the  last  term.  Tp  tiiis  quotient  add  the  last  term,  and  the  sum  will  be  the 
urn.  of  the  series. 

llulft  2  and  3,  are  demonstrated  umler  Theorehr  8  of  Geometrical  Proportion. 


242 


GEOMETRICAL  PROGRESSION. 


Rule  III. 
Or,  When  all  the  terms  are  given,  then,  from  the  product  of  the 
second  and  last  terms,  subtract  the  square  of  the  first  term  ;  this 
remainder  being  divided  by  the  second  term  less  the  ^rs^,  will  give 
the  sum  of  the  series. 

Examples. 
1.  If  the  series  be  2.  6.  18.  54.  162.  486.  1458.  4374.  what  is 
its  sum  total  ? 

First  Method. 
From  the  greatest  term  =  4374 
Subtract  the  least  =        2 

Pivid€  by  the  ratio,  less  1=3— 1=2)4372  diC  of  extremes. 


Quotienl=  2186 
Add  the  greater  extreme=  4374 


6560 


4374—2 

Or, h4374=6560  Ans. 

3—1 


Second  Method. 
Greatest  term=4374 
Multiply  by  the  ratio=       3 

Product=  13122 
Subtract  the  least  term=  2 

Divide  by  the  ratio,  less  by  1=3—1=2)13120 

6560  Ans. 

4374  X  3—2 

Or, =6560 

3—1 

Third  Method. 

Greatest  term=4374?^ 
Multiply  by  the  second  term=       6 

Product=26244i 
Subtract  the  square  of  the  first  term=2x2=         4 

Divide  the  remainder  by  the  2d.  term  less  the  lirst=6— 2=4)26240- 

Ans.     656^] 

4374x6—4 

Or, =6560., 

6—2 


GEOMETRICAL  PROGRESSION.  243 

2.  A  man  travelled  6  days,  the  first  day  he  went  4  miles,  and 
each  day  doubling  his  day's  travel,  his  last  day's  ride  was  128 
miles  ;  how  far  did  he  go  in  the  whole  ?  Ans.  252  miles. 

3.  A  gentleman,  dying,  left  5  fons,  to  whom  he  bequeathed  his 
estate  as  follows,  viz.  to  his  youngest  son  £1000  ;  to  the  eldest 
£5062  lOs  and  ordered  ,that  each  son  should  exceed  the  next 
younger  by  the  equal  ratio  of  1| ;  vyhat  did  the  several  legacies 
amount  to?  Ans.    £13187   lOg. 

PROBLEril  IV. 

Given  the  extremes  and  ratio,  to  find  the  number  of  terms. 

Rule. 
-   Divide  the  greatest  term  by  the  least ;  find  what  power  of  the 
ratio  is  equal  to  the  quotient,  then,  add  one  to  the  index  oi  that 
power,  and  the  sum  will  be  the  number  of  terms.* 

Examples. 
1.  If  the  least  term  be  %  the  greatest  term  4374,  and  the  raN'e 
3  ;  what  is  the  number  oi  terms  ? 
Divide  by  the  least  term=2)4374=grealest  term. 


3x3x3x3x3x3x3=quotient,  2187==7th  pow.  then  7H-1=8,  Ans. 
2.  A  gentleman  travelled  252  miles;  the  first,  day  he  rode  4 
miles  ;  the  last  128,  and  each  day's  journey  was  double  to  the  pre- 
ceding one  :     How  many  days  was  he  in  performing  the  journey  ? 

Ans.     6  days. 

Problem  V. 

Given  the  least  terniy  the  ratio,  and  the  sum  of  the  series,  to  find  the 

last  term. 

RuLK.  Multiply  the  sum  of  the  series  by  the  ratio  less  l,to  that 
f)roduct  add  the  first  term,  and  the  result,  divided  by  the  ratio,  will 
give  the  last  term.j 

Examples. 
1.  If  the  first  term  be  2,  the  ratio  3,  and  the  sum  of  the  series 
6560  :   What  is  the  last  term  ? 

*Lettheseriesbel,2,4,8,  lG,or,  1,  1x2,  1X22,  1x23,  1x24.     Divide  the 

1X24 
-greatest  terni  by  the  least,  and  =:24.     But  the  exponent  is  always  1  le^s 

than  the  number  of  terms,  whence  4-j- 1,  or  1  added  to  the  index  of  the  last  term 
will  give  the  number  of  term?.     Though  the  index  of  the  last  is  not  common  y 
given,  yet  as  the  quotient  arising  from  dividing  the  last  term  by  the  first  term,  is.     ^ 
always  some  power  of  the  ratio,  it  is  readily  found  by  multiplication,  as  in  the 
1st  Example. 

t  This  rule  follows  directly  from  Rule  2,  Prob.  III.  For,  as  the  product  of  the 
last  term  and  ratio,  diminished  by  the  first  term,  and  the  remainder  divided  by' 
tlie  ratio  diminished  by  1,  gives  the  sum  of  the  series ;  multiply  the  sum  of  the 
series  by  the  ratio  diminished  by  1,  to  the  product  add  the  fir;t  term,  and  di- 
T'ide  the  sum  by  the  ratio,  and  you  will  have  the  last  term. 


244  GEOMETRICAL  PROGRESSION. 

Sum  of  the  series=6560 
Multiply  by  the  ratio  less  1=       2 

Producl=13120 
Atld  the  least  tertn=         2 

Divide  their  sum  by  the  ratio=3)r3122 

3-^1x6560-1-2  ^  4374  Ans, 

Or, =4374  Ans. 

3 
2.  A  gentleman  performed  a  journey  of  262  miles  ;  the  first  day 
he  rode  4  miles,  and  each  day  after  the  first,  twice  so  far  as  the 
day  before  :  How  far  did  he  ride  the  last  day  ? 

Ans,  128  miles. 

Problem  VI. 

Given  the  least  term,  the  ratio,  and  the  sum  of  the  series,  to  Jind  the 

number  of  terms. 

Rule. 
To  the  product  of  the  sum  of  the  series,  and  the  ratio  minus  1, 
add  the  first  term  ;  which  sum,  divided  by  the  first  term,  will  give 
that  power  of  the  ratio  signified  by  the  number  of  terms.* 

Example. 
If  the  first  term  be  2,  the  ratio  3,  and  the  sum  of  the  series  80; 
What  is  the  number  of  terms? 

Sum=80 
Multiply  by  the  ratio  less  1=3 — 1=  2 

160 
Add  the  first  term=     2 


.  Divide  by  the  first  term=r2)162 

81  which,  found  in  the  Ta- 
ble of  Powers,  is  the  fourth  power  of  the  ratio,  therefore  the  num- 
ber of  terms  is  4. 

Problem  VII. 
Given  the  extremes,  and  the  su7n  of  the  sei'ies,  to  Jind  the  ratio. 
Rule.     From  the  sum  of  the  series  subtract  the  least  term  ;  di- 
vide the  remainder  by  the  sum  of  the  scries  minus  the  greatesi 
term,  and  the  quotient  will  be  the  ratio  t 

*  By  Prob.  II.  the  difference  between  1  and  that  power  of  the  ratio  indicated 
by  the  number  of  terms,  divided  by  the  ratio  less  1,  and  the  quotient  multiplied 
by  the  first  term,  gives  the  sum  of  the  series.  Whence,  if  the  sum  of  the  series  be 
multiplied  by  the  ratio  less  1,  and  the  product  be  added  to  the  least  term,  you 
will  have  the  product  of  the  first  term  and  that  power  of  the  ratio  signified  by 
the  number  of  terms.  Divide  then  the  former  product  by  tlie  least  term,  and  tbe 
quotient  will  be  that  power  of  the  ratio  signified  by  the  number  of  terms. 

t  This  rule  is  deduced  from  Rule  2,  Prob,  HI.  in  the  easiest  manner. 


m: 


GEOMETRICAL  PROGRESSION.  245 

Examples. 

1.  If  the  least  term  be  2,  the  greatest  term  4374,  and  the  sum  of 
the  series  6560  :     What  is  the  ratio  ? 

Sum  of  the  series=6560 
Subtract  the  least  term=       2 
Divide 


Ad& 


idethe  re.  by  thesumofthe  ?  =6560-4374=2 186)6Si?(3 
series,  mmus  greatest  term     y  ^  g^^r 

2.  A  debt  of  ^252  was  paid  in  Geometrical  Progression,  the  first 
payment  was  $4  and  the  last  gl28  :  In  what  ratio  did  the  payments 
exceed  each  other  ?  Ans.  2,  viz.  a  double  ratio. 

Problem  VIII. 

The  first  term,  the  number  of  terms^  and  the  last  term  given^  to  find 

the  ratio. 

Rule. 
Divide  the  greater  extreme  by  the  less,  and  extract  such  root 
of  the  quotient,  whose  index  is  equal  to  the  number  of  terms,  less  1. 
Or,  find  the  quotient  in  the  Table  of  Powers,  the  root  of  which  is 
the  answer.* 

Examples. 

1.  Given  the  extremes  2  and  4374,  and  the  number  of  terms  8  : 
What  is  the  ratio  ? 

Divide  by  the  least  term=2)4374=greatest  term. 

I  -^2187=3 


4374 
Or, 

2 


1 
=3,  Ans. 


Problem  IX. 
I'he  extremes  and  number  of  terms  given,  to  find  the  sum  of  the  series. 

Rule. 

1.  Subtract  the  least  term  from  the  greatest,  and  make  the  dif- 
ference a  dividend. 

2.  Divide  the  greatest  term  by  the  least,  and  extract  such  root 
of  the  quotient,  whose  index  is  equal  to  the  number  of  terms  less 
1  ;  take  1  from  the  said  root,  and  make  the  remainder  a  divisor. 
(Or  find  the  quotient  in  the  table  of  powers,  which  will  shew  the 
root,  from  which  subtract  1.) 

*  Let  the  series  be  1, 1x3,  1x32, 1  x33,  1 X34.    Divide  the  last  term  by 

1  X34 
the  least  term,  and  -. =34 ,  and  the  quotient  is  that  power  of  the  ratio,  whos» 

index  is  1  less  than  the  number  of  terms.     Extract  that  root  of  the  quotient, 
whos^  index  is  1  lecF  than  the  number  of  terms,  and  that  root  will  be  the  ratio. 


-MG 


GEOMETRICAL  PROGRESSION. 


3.  Divide  the  dividend  by  the  divisor,  and  the  greatest  term, 
added  to  the  quotient,  will  give  the  sum  of  the  series.* 

Examples  . 
Given  the  extremes  2  and  4374,  and  the  number  of  terms  8  : 
What  is  the  sum  of  the  series  ? 

From  the  greatest  term=4374 
Take  the  least=       2 


Blakc  this  remainder  a  dividend  4372 


Divide  the  greatest  term  by  the  least  2)4374 

7  

And  extract  the  7lh  root  of  the  quotient,  ^^2187=3  ;  Then» 

3—1=2)4372 


Q,uotient=2186 
A<Id  the  greatest  term =4374 


6560  Ans. 


4374--2 


Or,  4374- 


:6560 


4374 

2 


— 1 


Problem  X. 

Given  the  ratio,  the  number  of  terms.^  and  the  gteattit  term^  io  Jind 

the  least  term. 

Rule. 
Divide  the  greatest  term  by  such  power  of  the  ratio,  whose  in- 
{ie%  is  equal  to  the  number  of  terms  less  1,  and  the  quotient  will 
he  the  least  lerfu.j 

Example. 
If  the  ratio  be  2,  (he  number  of  terms  6,  and  the  greatest  terna^ 
128  :   What  is  the  least? 

Divide  the  last  term  by  2x2x2x2x2= 
power  of  the  ratio 

Or, ~4 


'^^1^32) 


128(4  Ans. 
128' 


*  This  rale  is  a  combmatloa  of  the  following;  steps.  Find  by  Frob.  VIII,  the 
fatio,  and  then  find  by  Note  1,  under  Prob.  I.  the  sum  of  tlic  series  exce])t  the 
4ast  terra,  by  diyidiiig  the  difference  of  the  extreme?  by  the  ratio  less  ] .  To  this 
ci^uotieat  add  the  last  term,  and  the  sum  will  evidently  be  the  sum  of  the  series. 

t  Let  the  geometrical  series  be,  2,  2  X  3,  2  X  33 ,  2  X 33 , 2  X  3 4 .     And  the  least 

terfflh,  2,  is  eviJejatly  ec^ual  to  the  last  term, =2.     As  the  index  of  tlie  last 

ffiwm  is  one  less  than  the  number  of  terms  ;  the  reason  of  the  rule  is  evident. 


iki 


GEOMETRICAL  PROGRESSION.  247 

Problem  XI. 

Qiven  the  ratioy  the  number  of  terms,  and  the  greatest  term^  to  find 
the  sum  of  the  series. 

Rule. 

1,  Divide  the  greatest  term  by  such  power  of  the  ratio,  whose 
index  is  equal  to  the  number  of  terms  less  1  :  take  the  quotient 
from  the  last  term,  and  make  the  remainder  a  dividend. 

2.  Divide  the  dividend  by  the  ratio  less  1,  and  the  quotieat, 
added  to  the  greatest  term,  will  give  the  sum  of  the  series.* 

Example. 
If  the  ratio  be  4,  the  number  of  terms  6,  and  the  greatest  term 
3072  :  What  is  the  sum  of  the  series  ? 

Divide  the  lastterm  by  the  }  =4x4x4x4x4=1024)3072(3 
51h  power  of  the  ratio     )        ^  ^  ^  ^  3072 


[ 


From  the  last  term=:3072 
Take  the  quotient=       3 

Divide  by  the  ratio  less  1=4—1=3)3069 

QuotieDt=1023 
Add  the  greatest  term=3072 

Ans.=4095 


3072 

3,072- 

46-1 

Or, 

30724 

, 

_ 

4095, 
4—1 

Problem  XII. 

Given  the  ratiOf  the  number  of  terms^  and  the  sum  of  the  series^  to 

find  the  least  term. 

Rule. 

Divide  the  ratio,  less  1,  by  such  power,  less  1,  of  the  ratio, 

whose  index  is  equal  to  the  number  of  terms,  and  the  quotient, 

multiplied  by  the  sum  of  the  series,  will  give  the  least  term.f 

Example. 
If  the  ratio  be  4,  the  number  of  terms  6,  and  the  sum  -of  the 
series  4095  :  What  is  the  least  term. 

*  Find  by  Prob.X.  the  least  term^and  then  find  by  Rule  1,  Prob.  III.  the  sum 
of  the  series.     This  process  corresponds  to  the  rule  in  the  text. 

t  By  Prob,  II.  the  ratio  is  raised  to  a  power  whose  index  is  tlie  number  of 
terms,  and  then  diminished  1,  and  the  remainder  is  divided  by  the  ratio  less  1. 
This  quotient  nmltiplied  by  the  least  term,  j^ives  the  sum  cf  the  series.  Whence 
the  least  term  must  be  equal  to  the  sum  cf  Uie  terms  divided  by  this  quotient, 
•-vhich  is  the  rnlo. 


m  GEOMETRICAL  PROGRESSION. 

4X4X4X4X4X4=4096,  and  4096—1=^4095,  then,  the  ratio  less  1, 

3  3        4095 

divided  by  4095,  is  j^,  and  J^X  ~Y~=3,  Ans. 

4—1 

Problem  XHI. 

Givtn  (he  ratio,  the  number  of  terms,  and  the  sum  of  the  swries,  to 

find  the  greatest  term. 

Rule. 

1.  Subtract  that  power  of  the  ratio,  which  is  equal  to  the  num- 
ber of  terms  less  1,  from  that  power  of  it,  which  is  equal  to  the 
whole  number  of  terms. 

2.  Divide  the  remainder  by  that  power  of  the  ratio  minus  uni-ty 
which  is  equal  to  the  number  of  terms,  and  the  quotient,  multipli- 
ed by  the  sum  of  the  series,  will  give  the  greatest  term.* 

EXAP4PLE. 

If  the  ratio  be  4,  tlie  number  of  terms  6,  and  the  sum  of  the 
s^eries  4095  :   What  is  the  greatest  term  ? 
From       4X4X4X4X4X4=4^=4096 
Subtract  4X4X4X4X4      =45  =  1024 

3072 

Divide  by  46—1=4095)3072= which    rmil^lied 

3072     4095  4095 

by  the  sum,  is x =3072  Ans. 

4095        1 
46 — 4.6-1 

Or,     46__i     X4095=3072 

3.  The  two  last  problems  may  be  solved  by  one  short  operation, 
thus  :  Divide  the  sum  by  the  ratio,  and  the  remainder  after  the 
operation  will  be  the  least  term  ;  then  take  the  quotient  from  the 
sum  of  the  series,  and  the  remainder  will  be  the  greatest  terra.j 

*  This  rule  is  a  contraction  of  the  following  process.     Find  by  Prob.  XII.  the 
least  term,  and  then  find  by  Prob.  I.  the  greatest  term.     Thus  in  the  exam- 
4—1  4—1 

pie  ; X4095=the  least  term  by  Prob.  Xll. ;  and  by  Prob.  I. X4095 

46—1  ,■  46—1 

4—1  4 — 1X45  46—45 

X46-i= X4095x4s=the  greatest  term= X4095= — : X 

46—1  46—1  46—1 

4095=3072,  the  greatest  term. 

t  Let  the  series  be  3+12-|-48-l-192-f  768+3072=409.p,  or  let  it  be  written 
3_|-3x4-f3x42+3x43+3x44-j-3X45=4095.     This  is  the  same  as  3-f3x 


4^-42 -f  43+44 -f  45  =4095,&itis  evidentthat4095-3=3  X4-{-42-|-43^-44 -i-4^ 


and  that  4095 — 3  contains  3x4-|-4a-j-43-j-44-f  45,  a  certain  number  of  times 

4095—3 
exactly.    If  then  both  be  divided  by  the  ratio  4,  we  shall  have =3X 


1-I-4-J-424-43-I-44— 3X341— 103.3=t.fee number  of  times  4  is  contained  exactly 


GEOMETRICAL  PROGRESSION.  249 

For  the  least  term.  For  the  greatest  term. 

4)4095(1023  quotient.  From   the   sum=4095 

4  Subtract  the  quolient=1023 

09  Ans.=3072 

8 

15 
12 

3  Ans. 

Problem  XlV^ 
Given  the  ratio,  ike  last  term^  and  the  sum  of  the  series  to  find  the 
first  term.  ' 

Rule. 
From  the  sum  of  the  series  take  the  last  term,  and  multiply  the 
remainder  by  the  ratio ;  then  take  this   product  from  the  sura  of 
the  series,  and  the  remainder  »vill  be  the  first  term.* 

Example. 
If  the  ratio  be  4,  the  last  terra  3072,  and  the  sum  of  the  series 
4095  ;  what  is  the  first  term  ? 

From  the  sum=4095 
Take  the  last  term=3072 

Remainder=1023 
Multiply  by  the  ratio=       4 

,  Subtract  4092  from  the  sum. 

And  the  remaimler         3  is  the  Answer. 

Problem  XV.  and  XVI. 

^ liven  the  numler  of  terms,  the  last  term,  and  the  sum  of  the  series,  to 
find  the  first  term  and  the  ratio. 
The  solution  of  these  two  Problems  being  very  tedious  by  the 
Theorems,  they  may  be  solved  by  a  very  short  operation  ;  thus, 

in  4095 — 3.  If,  however,  the  first  term  3,  be  not  taken  from  the  sum,  it  is  evi- 
dent that  the  ratio  must  be  contained  in  4095,  the  same  number  of  times,  or  10/23 
times,  with  a  remainder  always  equal  to  the  first  term,  which  is  in  this  exaropie 
3.     And  thus  also  for  any  other  case  wliere  the  first  term  is  less  than  the  ratio. 


Again,  3+3X4+42-f  434-44-|-45=4095,  or  as  it  may  also  be  written,  3>^ 


l-j-4-f42-f  43-1-44 +45  —  4095.    But,  we  have  seen  that  3XI+4-I-43-I-434-44 

4095—3  

is  contained  in just  1023  times,  and  3X  l+4-}-43 -{-43+44  is  the  series 

4    ' 
except  the  last  term  3  X  45 .     Therefore  4095—1023=3072,  the  last  term. 

*  This  rale  is  deduced  from  Rule  1,  Prob.  IJI.  and  will  he  evident  on  atti^ndi^^- 
to  that  rule . 

H  h 


f50 


GEOMETRICAL  PROGRESSION. 


Divide  the  sum  of  the  series  by  the  difference  between  the  sum 
and  the  last  term  ;  the  quotient  will  give  the  ratio,  and  the  remain- 
der, after  the  operation,  the  first  term.* 

Example. 

If  the  number  of  terms  be  4,  the  last  term  64,  and  the  sum  of 
the  series  80 ;  required  the  first  term  and  the  ratio  ? 

From  the  sum=80 
Take  the  last  term=64 

Divide  by  the  difference=26)80(3  the  ratio. 

78 

The  first  term=  2 


T/ie  following  Table  exhibits  a  summary  vierss  of  the  doctrine  of  Geo- 
metrical Progression. 


CASES  OF  GFOMETKICAL  f^ROGKESSlON. 

Cas"  1   Given    j    Pir.qaiivd  |                        Solutiot* 

1. 

am 

I 

7i— 1 

ar             Prob.  I. 

s 

n 

xa     Prob.  II. 

r— 1 

2. 

ad 

s 

I— a 

H Prob.  lU. 

r—l 

n 

LI— La 

.    .    .   I    1       Prnh     IV 

L.r 

*  By  the  second  rule,  Prob.  XIII.  it  is  found  that  if  the  sum  of  the  scries  be  di- 
vided by  the  ratio,  the  quotient  is  the  difference  between  the  sum  of  the  serie-j 
and  the  greatest  term,  and  the  remainder  is  the  first  term.  Therefore,  if  the  sum 
of  the  series  be  divided  by  the  difference  between  the  sum  and  the  last  term,  th^ 
quotient  must  be  the  ratio,  and  the  remainder  the  least  term,     hi  tlie  example 

4095 
there  taken,  the  ^catest  term  divided  by  the  ratio,  or  — — =:=1023  with  a  remam- 

der=3  the  first  term.     But  as  1023=4095—3072  or  the  difference  between  the 

4095 
sum  of  the  series  and  the  last  term,  therefore  -~: — —---=4,   the  ratio,   and 

leaves  a  remainder  3,r=thc  first  term.     The  same  must  hold  true  in  every  series, 
where  the  ratio  is  greater  than  the  least  term. 


GEOMETRICAL  PROGRESSION, 


251 


Cas-  1 

Giveu    1 

i.equiied 

Solntiofi. 

3. 

ars 

/ 

r— lX.s+« 

r             Prob.  V. 

n 

L.r—iXs-\-a — L.a 

L.r                 Prob.  VI. 

4. 

ah 

r 

s — a 

Prob.  VII.  . 

n 

L 

L.l-'L.a,^ 

.s — a — L.6 — / 

5. 

ans 

r 

rs 

11 
r 

a 

s — a 

a 

i 

I 

lXs—l\     =aXs—a\ 

6. 

anl 

r 

1 

a 

/I— 1     Prob.  VIII. 

s 

a 

/—a 

Prob.  IX. 

7. 

ml 

a 

I 

n—l     Prob.  X. 
r 

s 

H 

/ 

I, 

n-,1     Prob.  XI. 
r 

r-1 

8. 

1 
1 

rns 

a 

r— 1 

Xs    Prob.  XII. 
n 
r  —1 

I 

5 

71      n^ — 1 

•  •ys     Prob    XIII 

INTEREST. 


Case  I  Given    |  Required  |                        Solution.                      \ 

9. 

rh 

a 

s—rxs—i    Prob.  XIV. 

n 

L.I- 

I 

L.s — r  X  s — I 

1  1 

L.r            '^ 

10. 

nls 

a 

n—l in— 1         1 

=lxs—t\ 

aXs—a 

r 

n      s    n — 1         / 

J  1  ■        r          — 

l—s                I—s 

^  a=tir8t  or  least  term. 
1  /=Iast  or  greatest  term. 
TT        !s=sum  of  all  the  terms. 
^^^  }n=number  of  terms. 
1  r=ratio. 
J  L=lo^arithm. 

Examples  in  Geometrical  Progression, 

To   be  solved  by  those  formulas  in  the  Table,  of  which  the   rules 
are  not  given  in  the  Problems. 

1.  The  least  term  is  2,  the  greatest  term  4374,  and  the  sum  of 
the  series  6560  ;  required  the  number  of  terras.  Ans.  8. 

This  is  solved  by  the  2nd  rule  of  Case  4  in  the  table. 

2.  If  the  ratio  be  4,  the  last  term  3072,  and  the  sum  of  the  se- 
ries 4095  ;  what  is  the  first  term  and  the  number  of  terms  ? 

Ans,  The  first  term  3,  the  number  of  terms  6. 
Wrought  by  Case  9. 

3.  Given  the  number  of  terms  4,  the  last  term  54,  and  the  sum 
of  the  seVies  80,  to  find  the  first  term  and  ratio. 

Ans.  The  first  term  2,  the  ratio  3. 
Wrought  by  Case  9  in  the  table. 


INTEREST. 


INTEREST  is  a  premium  allowed  by  ihe  Borrower  to  the 
Lender,  for  the  n>e  of  his  nionev.  It  is  e.-timated  at  a  certain 
number  of  pounds,  dollars,  &,c.  for  each  huudrrd  pounds,  dollars, 
&:c.  for  a  year;  and  hi  ihh  same  proportion,  for  a  less  or  greater 
«um,  or  for  a  longer  or  &hoder  time.  He.ice,  interest  is  sa^id  lo 
be  ^0  much  jtjer  cent,  ov  per  centum ^  per  annum. 


SIMPLE  INTEREST.  5253 

The  Principal  is  the  sum  lent. 
The  Rate  is  the  premium  per  cent,  agreed  on.* 
The  Amount  is  the  sum  of  the  principal  and  interest. 
Interest  is  of  two  kinds,  Simple  and  Compound. 

SIMPLE  JJVTEREST. 

Simple  Interest  is  that  which  is  allowed  on  the  Principal  only. 

JVote.  By  this  Rule,  Commission,  Brokerage,  Insurance,  pur- 
chasing Stocks,  or  any  thing  else,  rated  at  so  much  per  cent,  ars 
calculated. 

General  Rule. 

1.  Multiply  the  Principal  by  the  Rate,  and  divide  by  100  (or  cut 
off  the  two  right  hand  figures  in  the  Pounds)  and  the  quotient,  or 
left  hand  figures  will  be  the  answer  in  Pounds,  &c.  the  right  hand 
figures  being  reduced  and  cut  off  as  at  first.  If  the  principal  b«* 
dollars,  the  right  hand  figures  will  be  cents.f 

2.  For  more  years  than  one,  muHiply  the  Interest  of  one  year 
by  the  number  of  years. 

3.  For  any  number  of  months  take  the  aliquot  parts  of  a  year ; 
and  for  days,  the  aliquot  parts  of  30. 

JVote.  When  ^  |  multiply  f  ^  |  of  the  given  number  of 
the  rate  percent.  <  .  [» the  prin- ^  ^  [.months,  and  you  will  have 
per  annum  is  o  j  cipal  by  |  J  |  the  interest   for  the    given 

Examples. 
1.  What  is  the  interest  of  £573  ISs.  9id.  at  G  per  cent  per 
innum  ?  Ans.  £34  8s.  5d. 

*  Laivful  or  legal  interest  is  that  whicli  is  permitted  by  the  laws  of  the  Slate. 
It  is  different  under  different  governments.  In  England  tlie  Rate  is^'t-e  perjceut. 
in  the  New  England  States,  it  is  six^  and,  in  the  State  of  New  York  it  is  seven 
per  cent.  The  courts  of  the  United  States  allow  interest  according  to  the  prac- 
tice of  the  State  where  the  suit  was  commenced.  The  rules  of  the  Courts  in 
the  States  of  Massachusetts,  Connecticut,  and  New  York,  for  computing  lesal 
interest,  will  be  given  immediately  before  Discount. 

t  This  rale  is  a  contraction  of  a  process  in  the  Rule  of  Three.     Thus  in  the 

£,  Rate.       Principal. 
14.  Example.     A',  100  ::  6  ::  £573  13s.  9^d. :    the  interest  for  a  year.    And 

— : ;~:.T~ — —-=£34  8s.  5d.  the  interest.     The  remainder  of  the  rule  is 

100 
«)V>vious. 

The  reason  of  the  rule  in  the  Note  above  is,  that  as  9  is  J  of  J  2  months,  if 
tl,c  rate  be  9  per  cent,  multiply  the  jjrincipal  by  f  of  the  months ;  if  the  rate 
'.e  8  per  cent,  by  -|  of  tl^o  niunber  of'  months  ;  and   if  6,  by  half  the  months, 


254 


SIMPLE  INTEREST. 


£573 

13 

H 

6 

£34-42 
20 

2 

9 

8-42 

12 

613 
4 

•52 
2.  What  is  the  interest  of  £329  17s.  6^d.  for  3  years,  7  months, 
and  12  days,  at  5  per  cent,  per  annum.  Ans.  £59  13s. 


£329  17  6id. 


16-49 
20 


7  81 


987 
.12 

10-52 
4 


6  mons. 

^ 

1  mon. 
10  days 

2  days 

Then, 

16  9  101  interest  of  1  year 
3. 


49  9  7|do.  of  3  years. 

8  4  111  do.  of        6  months 

1  7  5f  do.  of         1  month 

9  If  do.  of  10  days 

1  9f  do.  of  2  days 


Ans.  £59  13         do.  of  3y.  7m.  12d, 


210 


Or  thus  : 
£5 

6  months 


1  month 

10  days 

2  days 


£ 

s. 

d. 

j\ 

329 

17 

6| 

i 

16 

9 

3 

49 

9 

71 

i 

8 

4 

Hi 

1. 

3 

1 

7 

6f 

1 

9 

If 

1 

91 

£59  13  Ans. 

3.  What  is  the  interest  of  347  dollars  50  cen.ts,  at  G  per  cent. 
p^r  annum  for  a  year. 

p47-50     Or  ^100  :  6  ::  347.50c.  :  §20  85c. 
6 


20-8500     Ans.  J^20  85c. 
4.  What  is  the  interest  of  §797  13c.  at  6  per  cent,  per  annum, 
far  8  months  ? 


SIMPLE  INTEREST.  255 

§797-13 
4 


31-8852     Ans.  pi  S^c,  5j\m, 

5.  What  is  the  interest  of  §649  17c.  at  6  per  cent,  per  annuiHj 
for  15  months  ?  Ans.  g48  68c.  7fm. 

6.  Required  the  amount  of  £725  12s.  6d.  at  5  per  cent,  per 
annum,  for  a  year?  5=^j\\126  12  6 


36   6  n 


Ans.  £761   18   1^ 


7.  What  is  the  amount  of  §560  50c.  at  6  per  cent,  for  16  months.' 

§560  50 

8 


44-84 
660-50 

Ans.  §605-34c. 

8.  What  is  the  interest  of  §150  75c.  for  1  month,  at  6  per  cent, 
per  annum  ?  ijl50-75 

-75375     Ans.  75cts.  3f  mills. 
So  that  any  number  of  dollars,  considered  as  so  many  cents,  is 
the  interest  for  2  months,  at  6  per  cent,  and  the  half  of  it  is  the 
monthly  interest. 

coM^usslo^\  or  factorage. 

Is  an  allowance  of  so  much  per  cent,  to  a  Factor  or  Corres- 
pondent, for  buying  and  selling  goods. 

9.  Required  the  commission  on  £436  9s.  6d.  at  3i  per  cent, 

£436     9     6 
3X 


1309     8 
218     4 

6 
9 

1^-27  13 
20 

^ 

5-63 
12 

6-39 
4 

1-56     Ans.  £15  6  6}. 


1^> 


SIMFLE  INTEREST. 


to.  Kequireu  the  commission  on  ^649  75c.  at  1^  percent. 
649-75 
324*875 
162-4375 


11370625     Ans.  pi  37o.  Ofm. 

BROKERAGE 

h  an  alloTVance  of  so  much  per  cent,  to  a  person  called  a  Dro- 
ker,  for  assisting  merchants  in  purchasing  or  selling  goods. 

11.  Required  the  Brokerage  on  £911  12s.  at  5s.  or  J  percent. 
5s.=i|911    12  /     . 

2  27   IG 
20 

5-58      Ans.  £2  5s.  G^d. 
12 


6-96 
4 

3-84 
12.  Required  the  Brokerage  on  g876  21c.  at  33}  cents,  or  at 
percent.  i|876-21 


292-07     Ans.  $2  92c.  Of^m. 

BmiNG  AND  SELLING  STOCKS. 

Stock  is  a  general  name  for  the  capitals  of  trading  companies, 
or,  of  a  fund  established  hy  government,  the  value  of  which  is  va- 
riable 

13.  Required  the  amount  of  £375  15s  bank  stock,  at  £75  per 


cent  ? 


375  15 


187  17  6 
93  18  9 


Or   thus 
|25|i|375   In 
Subtract  93   18  9 


As  befora^£281   16  3 


Ans.  £281    16  3 
14.  Required  the  amount  of  ^2195  50c.  bank  stock,  at  125  per 
cent.  |25;i|21 95-50 

Add  548-875 


Ans.  ^2744-375 
15.  What  is  the  value  of  $6950  at  105  per  cent  ? 

"Ans.  jf;7297  50cts. 
46.  Value  of  £225  of  stock  at  95  per  cent.  ?    Ajis.  £213  15g. 


SIMPLE  INTEREST.  257 

TO  CALCULATE  INTEREST  FOR  DAYS, 

Rule  I. 

Multiply  the  principal  by  the  days,  and  that  product  by  the  rate, 
and  divide  the  last  product  by  365x100,* 

15.  Required  the  interest  of  £360  10s.  for  175  days,  at  6  per 
f:ent.  360-5x175x6 

100X365     =^10-^=^10  7s.  4|d. 

Rule  for  making  a  divisor  for  any  Rate. 

Multiply  365  by  100,  and  divide  by  the  rate.     Thus,  for  6  per 
365X100 

cent.      z =6083  divisor. 

o 

365X100  ^.  . 

For  5  per  cent. =7300  divisor. 

365X100 
For  7  per  Cent. ;:: =5214  divisor,  and   so  for  any  other 

rate.     Therefore, 

*  This  rule  is  the  result  of  a  statement  in  the  Double  Rule  of  Three,  as 
follows.  £  Rate;    PHn. 

As  100  :  6  ::  360-5  :  the  interest  for  1  year. 
Days.  Days. 

And  as  365  :       ::  175  :  the  interest  for  175  days.    Wrought  according 

to  the  Rule  we  have     ^ =£10  7s.  4 |d.  the  interest  required. 

365  X 1 00 
The  rule  for  finding  a  divisor  for  any  Rate,  is  a  contraction  of  this  result. 

Kor  -^^Q'^  X  17 J  X Q_3QQ.5     ^^^  ^  -— ^— -  .     But  dividing  both  parts  of  the 
365X100  365X100  "  ^ 

fi  fi  1 

last  fraction  by  6,  the  Rate,  we  have  -— — — ■=— — jrr.-;-  Therefore 360-5 

■^  36^X100     36500     6083 

a  ^.^        .«  1        360-5x175       ^      , 

X 175  X -=360-5  X 175  X  — -= .    In  tlie  same  way  divisors 

365X100  6083  60i'.3  ^ 

are  formed  for  any  other  rate.    Hence  too,  the  2d  Rule  is  obvious,  for ^' 

^  6083 

r=the  interest,  and  is  the  product  of  the  principal  and  days  divided  by  the  divi- 
sor formed  as  above; 

When  the  time  is  given  in  months.^  the  divisor  is  formed  in  a  similar  manner. 
Suppose  in  the  last  example  the  time  had  been  1 1  months.     Then, 
As  100  :  6  ::  360*5  :  the  interest  for  a  year,  and  as  12  :     ::  11  months :  the  inter- 
est for  11  months.     Then   --;^.  '^-  =  the  interest  =:  360*5  X 1 1 X — . 

152X100  1200 

^ — ■— .    If  the  Rate  were  5,  tlien  -~  —=:-—.     Hence  the  rule  is  evident. 

200  ISOO     240 

JVb/e.  As  365  days  :  5  per  cent.  ::  7300  days  :  100  per  cent.  And  as  12 
months  t  5  per  cent.  ::  240  months  :  190  per  cent.  Hence  it  is  evident  that  if 
the  Rate  be  5^  any  principal  will  gain  100  per  cent,  that  is,  will  double  in  7300 
days  or  240  months.  And  at  6  per  cent,  any  sum  will  double  in  6083  days,  or 
200  months,  ant  at  7  pfn-  cent,  in  5214^  days,  or  17 IC-  months. 

1    i 


35B^ 


SIMPLE  INTEREST. 


Rule  II. 

Multiply  the  princrpal  by  the  days  ;  divide  by  6083  for  6  per 
cent,  and  7300  for  5  per  cent,  (the  days  in  which  any  sum  will 
double  at  those  rates)  and  the  quotient  is  the  interest.  For 
months,  maltiply  the  principal  by  them,  and  divide  by  200  for  € 
per  cent,  or  240  for  5  per  cent,  (the  months  in  which  any  sum  will 
double  at  those  r;»tes)  and  the  quotient  is  the  answer. 

Hence,  when  interest  is  to  be  calculated  on  cash  accounts,  or 
accounts  current,  where  partial  payments  are  made,  or  partial 
debts  contracted;  multiply  the  several  balances  into  the  days  they 
are  at  interesty  which  should  be  done  at  every  transaction,  and  thf: 
sum  of  these  products  divided  by  6083  and  7300^  will  give  the  in 
tcrest  at  6  and  5  per  cent.  For  any  other  rate,  make  the  proper 
addition  or  deduction,  or  find  a  divisor  as  before  directed. 

When  partial  payments  are  made  at  short  periods,  subtract  the 
several  payments  from  the  original  suih  in  their  order,  placing 
their  dates  in  the  margin. 

16.  Suppose  abillofg35i>wasdae  January  1,  1807;  that  ^75  wa-* 
paid  February  3d,  ^50  March  5th,  $80  April  9th,  and  June  7th, 
^145 :  What  interest  is  due  at  5,  6  and  7  per  cent.  ? 


Dates.          1 

Bill. 

Days. 

Products. 

January  1. 
Feb.  3,  paid 

p50 

75 

33 
30 
35 
59 

11550 

Balance, 
March  5,  paid 

275 
50 

8250 

Balance, 
April  9,  paid 

Balance, 
■June  7,  paid 

225 
80 

145 
145 

7875 
8555 

7300)36230(4'963' 
Ans.  $4  96c.  3m.  at  5  per  cent. 

6083)36230(5-955 
Ans.  ^5  95ic.  at  6  per  cent. 

52  [4)36230(6 -948 
Ans.  $Q  94c.  8m.  at  7  per  cent. 


After  the  dates  are  placed  in  the  margin,  the  number  of  days  in 
each  of  those  periods  is  to  be  computed  and  marked  against  it3 
respective  sum  :  lastly,  divide  the  sum  of  the  products  by  6083,  &c. 

Interest  on  accounts  current  is  calculated  nearly  in  the  same 
manner. 

17.  Compute  the  interest  at  6  per  cent,  on  the  following  ac 
c<^unt  to  August  lOlh. 


SIMPLE  INTEREST.  258 

Dr.         Mr.  A.  Jones,  bis  account  current,  with  B.  Carr,        Or. 


1807. 

D. 

1807. 

D. 

Jan.  1,  To  Cash,     - 

-       560 

March  10,  By  Cash,     - 

-     120 

Feb.  10,  To  do. 

-    300 

April  25,  By  do.     - 

.,-     130 

May  15,  To  do.     - 

140 

June  16,  By  do. 

-     450 

July  25,  To  do. 

-     100 

July  SI,  By  do.     - 

^50 

1807. 

IDs. 

Days. 

1    Products 

1     Dr.     Cr. 

Jan.          Ij 

Dr. 

560 

"~4(r 

22400 

^560      120 

Feb.      10, 

Dr. 

300 

300      130 
140     450 

Dr. 

860 

28 

24080 

100      150 

March  10, 

Cr. 

120 

1100      850 

Dr. 

740 

46 

34040 

850 

April    25, 

Cr. 

130 

250  Balance. 

Dr 
Dr. 

610 
140 

90 

12200 

May      15, 

/i\j 

January    30 

Dr. 

750 

32 

24000 

February  23 
March       31 

June     16, 

Cr. 

450 

April         30 

Dr. 

300 

35 

10500 

May          31 
June         30 

July     21, 

Cr. 

150 

July          31 

Dr. 

150 

4 

600 

August      10 

July     25, 

Dr. 

Dr. 

100 

250 

16 

4000 

Days       221 

Aug.     10, 

0083)131820(21-67'2 
.Ans  g21  67c.  2m. 

221 

131820 

Here  the  sums  on  either  side  are  introduced  according  to  the 
order  of  their  dates  ;  those  on  the  Dr.  side  are  added  to  the  for- 
mer balance,  and  those  on  the  Cr.  side  subtracted.  Before  we 
i:alculate  the  days,  we  try  if  the  last  sum  ^250  be  equal  to  the  bal- 
ance of  the  account,  which  proves  the  additions  and  subtractions. 
And  before  multiplying  \ye  try  if  the  sum  of  the  column  of  days 
be  equal  to  the  number  of  days  from  January  1  to  August  10. 

When  payments  are  made  at  considerably  distant  periods,  it  is 
usual  to  calculate  the  interest  to  the  date  of  each  payment,  and 
add  it  to  the  principal,  and  then  subtract  the  payment  from  the 
.amount. 

18.  A  note  was  given  for  ^540  the  18th  August,  1804,  and  there 
was  paid  the  19th  of  March,  1805,  g.50,  and  the  19th  of  December, 
1805,  $25  ;  and  the  23d  of  September,  180G,  |^25  ;  and  the  18th 
of  August,  1807,  gllO:  Required  the  interest,  and  balance  due  on 
.he  '.  Iih  of  November,  1807,  at  ^  per  cent.  ? 


^  blMPLE  INTEREbl. 

A  note  given  18th  August,  1804,  for 

interest  to  19th  March,  1805,  218  days,  gl9-35 


Paid  19th  March,  1805, 

Balance  due  19th  March,  1805, 
Interest  to  I9th  Dec.  1805,  276  days, 
Balance  due  19th  Dec.  1805, 
Paid  19th  Dec.  1805, 

Balance  due  19th  Dec.  1805, 
Interest  to  23d.  Sept.  1806,  278  days, 

Balance  due  23d  Sept.  1806, 
Paid  23d  Sept.  1806, 

Balance  due  23d  Sept.  1806, 
Interest  to  18th  Aug.  1807,  329.  days, 

Balance  due  18th  Aug.  1807, 
Paid  18th  Aug.  1807, 

Balance  due  18th  Aug.  1807, 
Interest  to  11th  Nov.  1807,  85  daya, 

Balance  due  11th  Nov.  1807, 

Amount  of  interest  paid,  ^98-823 

19.  A  owes  B  the  following  sum?,  wi(h  interest  at  6  per  cent, 
per  annum  :  $60  for  7  months,  gl50  for  9  months,  J75-50  for  S 
months,  g365-25  for  8  months,  and  510-20  for  5  months :  Kef^'ui-cd 
the  amount? 

g     60      X7=:  420 

150      X9=1350 

75-50x3=  226-^0 

365-23x8=2922 

510-20X5=2551 

1160-95  200)7469-50(37«347  Interest. 
1160-95    Principal. 


19-352 

;Jo40 
19-352 

559-352 
50 

23-022 

609-352 
23-022 

532-374 
2500 

23-197 

507-374 

23-197 

530-571 
25000 

27343 

505  571 
27-343 

532-914 
110- 

5-909 

422914 
5-909 

428-823 

Ans.  §1198  297  Amount. 

20.  A  note  for  glOOO  is  given  January  1,  1803,  with  interest  at 
6  per  cent  per  annum  ;  February  19,  1803,  glOO  are  paid  ;  June 
7,  1803,  giaO;  April  14,  1804,  g37-50;  July  U,  1804,  §76;  Sept. 
29,  1804,  ^,250;  Dec.  17,  1805,  g39  ;  March  4,  1806,  ^175; 
Aug.  7,  1806,  jil05;  Oct.  30,  1806,  g50 ;  May  12,  1807,  gS40,  and 
Nov.  17, 1807,  g72  :  How  much  is  due,  January  1,  1808  't 


SIMPLE  INTEREST  BY  DECIMALS. 


26  J 


SIMFLE  INTEREST  BY  DECIMALS. 

A  TABLE  OF  RATIOS,  FROM  ONE  POUND,  &C.  TO  TEN  POUNDS. 


j  Kate  per  cent. 

ratios. 

rate  per  cent. 

ratios. 

rate  per  cent. 

ratios. 

1 

'01 

4 

•04 

7 

•07 

H 

0125 

n 

•0425 

^ 

•0725 

H 

•015 

H 

•045 

^\ 

•075 

n 

•0175 

4 

•0475 

n 

•0775 

2 

•02 

5' 

•05 

8 

•08 

n 

•0225 

H 

•0525 

8i 

•0825 

f             2i 

•025 

^ 

•055 

8i 

•085 

2£ 

•0275 

5} 

•0575 

H 

•0875 

3 

•03 

6 

•06 

9 

•09 

3} 

•0325 

6^ 

•0625 

9i 

•0925 

3^ 

•035 

H 

•065 

9i 

•095 

3^ 

•0376 

6i 

•0675 

9^ 

•0975 

10      I  1 


Ratio 


.*-L.«  .^  the  Simple  Interest  of  £l  or  ^1  for  1  year,  at  the  rate 
per  cent,  agreed  on,  and  is  found  by  dividing  the  rate  by  100,  and 
reducing  it  to  a  decimal.     Thus,  7^0  =  ^^»  *'^^»  rf  o="05,  and  so  on. 

A  TABLE  for  the  ready  finding  of  the  decimal  parts  of  a  year, 
equal  to  any  number  of  days,  or  quarters  of  a  year. 


l^ays.   I  decimal  parts.  |     days.     |  decimal  parts.  |    days.   |    dec.  parts,    j 


•00274 

•005479 

•008219 

•010959 

•013699 

•016438 

•019178 

•021918 

•024657 


10 
20 
30 
40 
50 
60 
70 
80 
90 


•027397 
•054794 
•082192 
•109589 
•136986 
•164383 
•191781 
•219178 
•246575 


100 
200 
300 
365 


•273973 

•547945 

•821918 

1  -000000 


I  ol  a  yeai=  25 
J- of 


a  vear: 


^  of  a  year=  7; 


CASE  I.* 

Tlie  principal,  time,  and  ratio  given,  to  find  the  interest  and  amount. 

Rule. 
Multiply  the  principal,  time  and  ratio  continually  together,  and 
the  last  product  will  be  the  interest,  commission,  brokerage,  &c.  to 
which  add  the  principal,  and  the  sum  will  be  the  amount. 


*  This  is  a  contraction  of  the  Geneml  Rule  for  Simple  Interest.     If  the  iuter- 
ffst  on  £30  or  $30  was  required  for  2  years  at  6  per  cent,  by  the  general  rule, 

the  interest  is  -^-rr^XS— 30x^:^^X2— 30  X^ 06X2,  which  is  the  product  of 


100 

ivrincipal,  ratio,  and  time 


100 


And  the  amount=30+30X'06X2=je33^6  or$. 


m^  SlMi^LE  INTEREST  BY  PECIMALS. 

Examples. 
1.  Required  tbe  amount  of £537  10s.  at£6  per  cent,  per  annum^ 
fi>f  5  j'cars  ? 

Principal  537-5 
Multiply  by  the  ratio=     -06 

Product  32-250 
Multiply  by  the  time=^  5 

.    In(erest=161-250 
Add  tbe  piincipal=537-5 

Amount=698-75     ' 
20 


_. „ 1500  Ans. £698   159, 

Or,  537-5X06x5-f537-5— £698   15s. 

2.  What  is  the  simple  interest  of  £917  16s.  at  £5  per  cent,  ppk- 
^nnnom,  for  7  years  ?  Ans.£321   4  7. 

S.  What  is  the  amount  of£391  17s.  at£4|  per  cent,  per  annum, 
for  3}  years  ?  Ans.  £449  3   If 

4.  What  is  the  amount  of  £235  3s.  9d.  at£5|  per  cent,  per  an- 
mim,  from  March  5th,  1784,  to  Nov.  23d,  1784  ? 

Ans.  £244  0  81. 

5.  If  my  correspondent  is  to  ha?e£2|  per  cent ;  what  will" his 
commission  on  £785  13s.  amount  to  ?  Ans.  £19  12  10|. 

6.  What  will  be  the  interest  and  amount  of £445  lOs.  in  3  years 
and  129  days,  at  £8-}  per  cent,  per  annum  ? 

Ans.  Interest, £126  19  8i,  and  the  amount=£572  9  8i. 

7.  If  a  broker  disposes  of  a  cargo  for  me,  to  the  amount  of.t!i)o7 
lO.-i.  on  commission  at£li  per  cent,  and  procures  me  anothercargo 
of  the  value  £817  15s.  on  commission  at£l|  per  cent.  ;  what  wiii 
liis  commission,  on  both  cargoes,  amount  to  ?         Ans.  £22  5  7. 

8.  What  is  the  simple  interest  of  $^6-666  for  1|  years  at  7  per 
^ent.  ?  Ans. ^8   16e.  6m. 

9.  Find  the  amount  of  §1  for  9  years  s^nd  200  days,  computing 
interest  at  7  per  cent.  ?  Ans.  $1  66c.  8m. 

10.  What  is  the  interest  of  jj23G  at  5  per  cent,  for  one  year  and 
MO  days  ? 

11.  lloquired  the  interest  on  g648a  at  6  percent,  for  two  year§, 
six  months  and  20  days. 

CASE  II. 

T/te  amount,  titnc,  and  ratio  given,  iojind  the  principal. 
Rule. 
MuhipTy  the  ratio  by  the  time  ;  add   unity  to  the  product  for  a 
divisor,  by  which  sum  divide  the  amount,  and  the  quotient  will  be 
the  principal.* 

*  In  the  demonstration  of  the  R-iilr  A.r  Cji.«e  I.  it  was  proved  thutlhc  amonrl 
r^^ic  principal  added  to  tho  product  ofthe  principal,  ratio,  and  time,  or.  tsikii  if 


SIMPLE  INTEREST  BY  DECIMALS.  ms 

Examples. 

1.  What  principal  will  amount  to £1045  14s.  in  7  years,  at  £^ 
per  cent,  per  annum  ? 

Ratio=^-OD 
Multiply  by  the  time=:     7 

Product=-42 
Add  1- 

Divisor==l-42)lO45'7(736.40844-— £736  8  2, 
1045-7 
Or, =£736  8  2  Ans. 

•06x74-1 

2.  What  principal  will  amount  to  £3810,  in  6  years,  at  £4i  per 
cent,  per  annum  ?  Ans.  £3000^ 

3.  What   principal  will  amount  to  £^}66  Ps-  0^-  in  3^  years^  at 
£  f)l  per  cent,  per  annum  ?  Ans.  £563. . 

4.  What  principal  will  amount  to  £335  7s.  3d.  in  3  years  and  9T 
days,  at  £9^  per  cent,  per  annum  ?  Ans.  £255   ID  0-\ 

CASE  HI. 

The  amount  J  principal,  and  time  given,  to  find  the  ratio. 

Rule. 
Subtract  the  principal  from  the  amount;  divide  the  remainder 
hy  the  product  of  the  time  and  principal,  and  the  quotient  will  be 
the  ratio.* 

Examples. 
I.  At  what   rate  per  cent,  will £543  amount  to £705  I8«.  in  ^'■■ 
rears  »  l^'rora  the  amount=705-9 

Take  the  principal=543' 


Divide  by  543x5=^27 15)1 62. 90(0^ 
162-90 


Ike  same  example,  the  amount,  33'6=30-f  30X*0t>X2,  or  which  is  the  same 
};hing,= 1  -f  •()(>  X  !^  X  30.     Divide  both  by  the  same  quantity^  1  +  -Ob  X  2,  and  the 

•           ^^^    .-11  u             111                ^'^          rPoGxixSG 
expression  will  still  be  equal,  and  we  have  — ; — ; — — = ;     thca 

cancel  the  equal  terms  in  the  last  fraction,  and  — ; ■ — =30,  that   is,  the  a- 

mount  divided  by  the  product  of  the  ratio  and  time  increased  by  1,  gives  a  quo- 
tient, which  is  the  principal.  The  same  may  be  shown  in  any  other  example, 
and,  hence  the  rule  is  general. 


*  Under  case  I.  it  was  shown  that  the  amount,  33'ar=:30+3U  X  "UO  X ii.  T-dk« 
the  principal,  30,  from  both  sides,  and  33-G— 30r=z30X  '06x2,  or  3-6=:30X2X'06. 
_    .      ,     ,  ,3-6        30x2X.0e 

Divide  both  parts  by  the  product  of  time  and  principal,  and  -— — r,=n77CTi — ^ — 

or  — =-06,  the  ratio,  and  illustrates  the  rule. 

30X2 


364  SLUPLE  INTEREST  BY  DECfMALS. 

705-9— 543 

.    Or, =-OG~£GAns. 

543x5 

2.  At  what  rate  per  cent,  will  £391  17s.  amount  to  £449  os.  l|d, 
74qr.  in  3}  years  ?  Ans.  £4^.. 

3.  At  what  rate  per  cent,  will  £4 13  12s.  6(1.  amount  to  £546  4s. 
lt)id.  in  4|  years  ?  Ans.  £6^. 

4.  At  what  rate  percent.  will£3000  amount  to  £3810  in  6  years? 

Ans.  £4|. 

CASE  IV. 

The  amounty  principal^  and  rate  per  cent,  giverty  to  Jind  ike  time' 

Rule. 

Subtract  the  principal  from  the  amount ;  divide  the  remainder 
by  the  product  of  the  ratio  and  principal ;  and  the  quotient  will 
he.  the  time.* 

Examples. 

1.  In  what  time  will  £543  amount  to  £705  IGs.  at  £6  per  cent- 
ner annum  ? 

From  the  amount=705-9 
Take  the  principal=543 

Divide  by  543X-06=:32-58)  162-9(5  years,  Ans. 

162  9 

2.  In  what  lime  will  £3000  amount  to  £3810,  at  ^  per  cent, 
per  annum  ?  Ans.  6  years. 

3.  In  what  lime  will  £391  17s.  amount  to  £449  3s.  l|d.  at£4i 
per  cent,  per  annum  ?  Ans.  3}  years. 

7b  ^rind  the  Interest  of  amj  Sum^  at  6  per  cent,  per  annum,  for  any 
number  of  months. 

Rule. 

If  the  months  be  an  even  number,  multiply  the  principal  by  half 
that  number  ;  and  if  the  months  be  uneven,  halve  the  even  months, 
to  which  annex  /^  ;  thus  the  half  of  19  is  9  5  ;  and  multiply  the 
principal  as  before,  dividing  by  100  or  cutting  oft'  two  figures  more 
at  the  right  hand,  than  there  are  decimals  in  both  factors,  which 
reduce  to  farthings,  each  time  cutting  oft'  as  at  first. 

4.  What  is  the  interest  of  £345  16s.  6d.  for  9  years  and  11 
months,  at  6  per  cent,  per  annum  ? 

.^     ^-  3-6        30X'06X2   '  3-fi    '        ^^    ;.  _         .„    ^ 

*    Also,  =- — ' ,  and -=2,  the  time,  and  is  an  ilUustj.T- 

30X-0«        30X-0U  mx'06 

tion  of  Ihc  Rule  for  Case  IV. 


SIMPLE  INTEREST  BY  DECIMALS. 


9.&i 


y.  m. 

9  11 
12 

345-825     2)119  months. 

59-5 

59'5=i  No.  of  months. 


1729125 
3112425 
1729125 


je205'765875=:  £205  15  3^  Ans. 
Principal='je345  16  6 


Amount: 


£551   11  92 


A  Table  of  decimal  parts  for  every  day  in  the  twelfth  part  of  a  year, 
which  consists  of  36  5^  days. 


1  dai/s.  \dec.  pts.\  days.  \d€c.  pts.\  days. 

dec.  pts.\  daifs. 

dec.  pts.\  days.  \dtc.  pts.\ 

1 

•033 

7 

•230 

13 

A21 

19 

•624 

25 

•821 

2 

•066 

8 

•263 

14 

•460 

20 

•657 

26 

•854 

3 

•098 

9 

•296 

15 

•493 

21 

•690 

27 

•887 

4 

•131 

10 

•328 

16 

•526 

22 

•723 

28 

•920 

5 

•164 

11 

•361 

17 

•558 

23 

•756 

29 

•953 

6 

•197 

12 

•394 

18 

•591 

24 

•788 

30 

•986 

To  find  the  Interest  of  any  Sum,  either  for  Months,  or  Months  and 
Days  at  6  per  cent,  per  annum. 

Rule. 
Multiply  the  principal  by  the  number  of  months,  (or  months  and 
parts,  answering  to  the  ^\vea  number  of  days  in  the  table)  and  cut 
off  one  tig'ure  at  the  right  hand  of  the  product  more  than  is  re- 
quired by  the  rule  in  decimals,  and  the  product  will  be  the  inter- 
est for  the  given  time,  in  shillings  and  decimal  parts  of  a  shilling.* 


*  In  the  .iVb/e,  under  the  General  Rule  for  Simple  Interest,  it  is  shown  that 
■v^'hen  the  Rate  is  6  per  cent,  the  product  of  the  principal  and  ?ial/  the  number 
of  months  divided  by  100,  ^ives  the  Interest*  Whence,  the  product  of  the  prin- 
cipal and  the  number  of  months  divided  by  100,  must  give  twice  the  Interest. 

30*5  X 1 7 
Let  then  the  principal  £30*5  be  put  to  interest  for  17  months.     Then  — 

— £5*185=2X  £2-5925=twice  the  interest,  and  the  interest  is  2*5925.     Multi- 
ply by  20  and  the  interest  will  be  reduced  to  shillings  and  the  decimal  parts  ol' 
30'5  X  1 7  X  '^0 

a  shilling,  and  we  have   — -"— =2x£2-5925X20.      Divide  by  2,  and 

30-5X17X10  ^^ 

' Tm =£2-5925X20,  and  dividing  both  parts  of  the  fraction  by  10, 

30-5  X  17 


10 


— £2-5925  X  20,  that  is,  multiply  the  principal  by  the  number  of  months 

■^ni\  divide  the  product  by  10,  or,  cut  off  only  one  figure  more  than  the  rule  for 
K  k 


ii)60,  SIMPLE  INTEREST  BY  DEGIMALSv 

Examples. 

1.    What   is   the    inlercst  of       2.    What  is   the    interest   of 

£  100,  for  a  year  ?  £  250  10s.  for  19  months  and  7 

Principal=100  dajs  ? 

Blult.by  the  months=12  Principal=  £250-5 


Ans.  s.l20|0=£6 


Time=     19-23 


7615 

JVote.  This  Table  may  also  be  5010 

tised  for  the  parts  of  a  year^  in  22545 

Compound  Interest,  after  having  2505 

Worked  for  whole  years.     The  

shillings,  &c.  in  the  principal  must  Ans.  s. 481-7115 

iirst  be  reduced  to  the  decimal  —  £24  1  8^ 
of  a  pound. 

Another  Method  of  calculating  lutenst  for  Months,  at  £6  per  cent, 

psr  annum. 

RtJLE. 

If  the  principal  consists  of  pounds  onfy,  cut  off  the  unit  figure, 
and,  as  it  then  stands,  it  will  be  the  interest  for  one  month  in  shil- 
lings and  decimal  parts  : — If  it  consist  of  pounds,  shillings,  &;c.  re- 
duce the  shillings,  &c.  to  decimals,  which,  with  the  unit  figure  of 
the  pounds,  will  be  decimal  parts  of  a  shillmg.* 

Examples. 

1.     What  is  the   interest   of  2.     What  is   the   interest   of 

£  175,  for  5  months  ?  £255  I6s.  for  7  months  ? 

£175=17-5  shill.=interest  for  Shill. 

1  month,                              17-5  £255  16=25-58  int.  for  1  mr. 

Multiply  by  the  time=      5  7 

2|0)87-5  2|0)179  06 

An&.=£4  7  6  Ans.  £€  ^19  0| 

decimals  requires,  and  you  have  the  inlerest  in  shillings  and  decimal  parts  of  a 
shilling.  If  there  be  months  and  days,  the  days  being  made  decimals  from  the 
table,  the  same  rule  would  manifestly  apply. 

*  This  rule  is  only  a  contraction  of  the  following  process  by  the  Double  Rule 
of  Three,  to  find  the  interest  of  any  sum,  e.  g,  £36,  for  1  month. 
As  100  :  6  ::  36 

12  :       ::  1  month  :  the  interest. 

6  X  36  6  X  36  X  '^O 

Hence  -'— — ^^=^£*18=:3'6s.  the  interest  for  1  montli.     But ^^ ~ 

12X100  ...  •  1-2X100     "" 

the  interest  in  shillings,  or  shillings  and  decimals,  and  cancelling  the  equal  pari? 
6X36X20     1X36X2     1X36X1      36     ^  ^   ,  .„.  '    ,  ,         ,      , 

^^  *^"^'  •iisaob-"=^-237ir  =-13^0-  ="1^=^^*^  '^^^'^-''  ^' ^^"^^  ^' ''': 

cimals  of  a  pound,  the  rule  would  be  equally  correct,r 


SIMPLE  INTEREST  IN  FEDERAL  MPNEY.         2G7 

SIMPl^E  INTEREST  IN  FEDERAL  MONEY. 

Pboblem  I. 
When  the  principal  is  given  in  JWw  England  pounds^  shillings,  ^-c. 
and  the  anni^iil  interest  is  required  in  federal  money,  at  6  per  cent. 

Rule. 
Reduce  the  shillings,  &c.  to  their  equivalent  decimal,  hy  inspec- 
iion,  divide  the  whole  by  5,  and  the  quotient  is  the  annual  inter- 
est :  Or,  multiply  the  principal  by  2,  and  the  product  (having  the 
tinlt  figure  of  the  pounds  cut  off)  will  be  the  interest  as  before.* 

Examples. 
I .  Required  the  annual  interest  of  £517  3s.  74d.  at  6  i>€r  cent.  ? 
3s.  =  -15  6)517181 

7id.=  030  D.     cm. 

Excess  of  12=001  103-436=103  43  6  Ans. 

.  Or,  517181 

•181  2 

D.    c.    m. 

103-4362=103  43  6y\. 
^.  Required  the  annual  interest  of  £1,  in  cents  ? 

Ans.  20  cents. 

Problem  11. 

When  the  principal  i^  given  in  New  England  currency,  and.  the  in- 
terest  and  amount  are  required  in  federal  money  at  6  per  cent. 

Rule. 
Reduce  the  New  England  money  to  federal,  then  divide  the 
principal  by  20  and  that  quotient  by  5 ;  add  those  quotients  to- 
gether, and  they  are  the  interest;  or  add  thetn  to  the  principal, 
did  their  sum  is  the  amount. t 

Examples. 
1.  Required  the  s^mount  of  £425  I63.  8?,d.  for  1  year,  at  6  per 
•ent.  ? 

*  This  rule  is  a  contraction  of  the  following  process.     By  the  General  Rulf; 

for  Simple  Interest,  (in   the  first  example)  the  annual  interest^:" — — — 

100 
This,  multiplied  by  20,  would  be  reduced  to  shillings  and  deciiAals  of  a  shillin;^, 
•and  divided  by  ^,  would  be  reduced  to  dollars  and  decimals  of  a  dollar.     Then, 

.M728LX6x50^51!:l£L=li=il!l»i  =  iJ,03  43c.  6^.^™. 
lOOXti  5X1  5  *  '« 

100X6  10X1  10  * 

+  Dividing  the  principal  by  20,  gives  the  interest  at  5  per  cent,  because  5  is 
>ae  twentieth  of  a  hundred;  then  dividing  this  quotient  by  5,  evidently  gives 
■«  he  interest  for  1  per  cent.  Then,  as  5-{- 1=6,  the  sum  of  the  two  quotients  will 
be  the  interest  at  6  per  cent.  ■  '      - 

Interest  at  6  per  cent,  may  often  be  calculated  most  easily,  by  finding  the  in- 
terest at  5  per  c^nt.  and  adding  one  fifth  of  this  interest  to  itself  for  6  per  cent. 
And  a'id  two  fifihs  of  it  to  itself,  and  vmi  Viave  tho  interest  at  7  per  cent. 


268       INTEREST  ON  BONDS  AND  OBLIGATIONS. 


•8 

3)425-835 

•034 

20)1419-460 

•001 

5)70-9725 

•835 

H1945 

D.     c.  m. 
1504-6170=1504-61   7.  Ans. 
2.  Required  the  amount  of  £112  4s.  6d.  for  one  year  ? 

Ans.  g396'52c  7io.  9dec. 

Problem  III. 

When  the  principal  is  JVew  England  currency,  and  the  monthly  r?j- 
terest  is  required  in  federal  money. 

Rule. 

Reduce  the  shillings,  &c.  to  decimals,  by  inspection,  then  sepa- 
rate the  right  han«l  figure  of  the  pounds  with  the  decimals,  divide 
by  6,  and  the  quotient  is  the  answer  in  dollars,  cents,  &c.* 

Example. 
Required  the  monthly  interest  of  £425  16s.  S^d.  in  federal  mo- 
ney? •  -8 

•034  6)42-5835 

•001  D.  c.  m. 

709725==7'09  71  Ans. 

£•835  •     . 

INTEREST  ON  BONDS  AND  OBLIGATIONS, 

HAVING  PARTIAL  PAYMENTS  ENDORSED. 

As  there  is  much  diversity  of  opinion  relative  to  the  computation 
Q>{  laxsoful  interest  in  these  cases,  several  Rules  will  be  given,  con- 
nected with  the  practice  of  the  Courts  of  several  of  the  States. 
The  difference  of  the  rules  depends  on  the  principle  assumed  in 
respect  to  the  time  Zi'lien  interest  becomes  due. 

Rule  L 

The  foUovving  rule  is  generally  thought  to  allow  too  little  inter- 
est.    It  is,  however,  adopted  in  some  of  the  Slates. 

Find  the  amount  of  the  principal  and  interest  for  the  whole  time  ; 
then  find  also  the  amount  q(  each  endorsement  for  the  time  it  has 
been  paid.  From  the  lirst  amount  deduct  the  sum  of  the  amounts 
of  thr;  several  endorsements,  and  the  remainder  is  the  balance  due. 

This  rule  involves  the  following  points; 

1.  Interest  is  not  due  until  the  obligation  is  paid. 

^42.vrCiX20 

•   This  rule  is  a  contraciion  of  the  foilowinjf  process.  * ^—^=:  tliR 

6 

..,.,,,       ^,       425-r.n5y?0Xf)     4e'5-8n5XlXl     4^vte     A'^-^P.SS 

r-nncipal  in  clOii?.     I  hen ■ ~ — == — — =r t=:-.-- —  i 

^'    .   ^  6XKHJXl'i  1X10X6  lUXt.  C 

th6  "mpnlhly  interest  In  ilollars  and  decimals  of  a  dcllar. 


INTEREST  ON  BONDS  AND  OBLIGATIONS.       iLC9 


* 


2.  Hence,  interest  must  be  allowed  on  the  endorsements  from 
*he  time  they  were  severally  made. 

Note.  A  shorter  method  of  computing  interest  according  to  this 
rule,  may  be  seen  in  examples  16  and  17  of  Simple  Interest.  The 
absurdity  of  this  rule  may  be  seen  in  the  following  manner.  Sup- 
pose I  borrow  glOO  at  6  per  cent,  for  ten  years,  and  pay  six  dol- 
lars at  the  end  of  each  year,  what  will  be  due  at  the  end  of  10 
years  ?  The  amount  of  glOO  is  gl60.  But  the  first  endorsement 
of  g6  has  borne  interest  for  9  years  ;  the  second,  for  8  years  ;  the 
third,  for  7  years,  and  so  on  ;  so  that  six  dollars  have  in  fact  been 
drawing  interest  for  forty  Jive  years,  and  thus  produced  g  16-20  of 
interest.  This,  added  to  the  nine  endorsements  of  §6  each,  gives 
g70-20.  That  is,  while  I  have  paid  only  the  annual  interest  of  §6. 
the 'principal  has  actually  been  reduced  §16  20.  By  paying  J6 
annually  for  25  years,  and  computing  interest  on  the  several  en- 
dorsements by  this  rule,  the  whole  principal  would  be  paid,  and 
the  lender  would  be  indebted  to  the  borrower  the  sum  of  §2,  while 
in  fact  the  lender  had  received  no  part  of  the  sura  lent. 

Rule  II. 

The  following  rule  was  drafted  by  the  late  Judge  Sedgwick,  and 
ordered  by  the  Court  of  Com.Tion  Pleas  for  the  County  of  Berk- 
shire in  1791,  and  is  the  Rule  now  used  by  the  Courts  of  Massa- 
'chusetts. 

**  On  all  contracts  carrying  interest,  on  which  partial  payments 
may  have  been  made,  the  principal  sum  with  the  interest  thereof 
shall  be  formed,  at  the  time  of  payment,  into  an  aggregate  sum, 
from  which  shall  be  deducted  the  sum  paid,  if  one  year's  interest 
shall  have  become  due,  and  if  not,  then  the  interest  shall  be  cast 
to  the  end  of  the  year,  and  the  aggregate  formed  as  aforesaid,  and 
from  the  same  the  payment  or  payments  and  the  interest  thereof 
from  the  time  or  times  of  payment  shall  be  deducted,  and  the  bal- 
ance or  balances  thus  formed  shall  bear  interest,  and  so  from  time 
to  time,  provided,  that  in  »J0  instance^  interest  shall  be  cast  on  a 
greater  sum  than  the  principal  sum  nor  on  any  interest  accrued." 

[Records  of  the  Court.] 

This  rule  involves  the  following  principles. 

1.  That,  when  an  obligation  has  borne  interest  for  one  or  more 
years,  interest  is  not  due  at  intervals  of  time  less  than  one  year. 

2.  Interest  is  to  be  computed  to  that  endorsement,  which,  to- 
gether with  the  preceding  endorsement  or  endorsements  and  its 
or  their  interest  since  the  time  of  payment,  shall  be  equal  to  or 
<^'xceed  the  interest  on  the  principal  when  this  endorsement  is 
made,  provided  one  year's  interest  shall  have  accrued.  The  re- 
mainder is  a  new  principal,  on  which  interest  is  to  be  computed 
as  before. 

3.  Interest  is  allowed  on  all  endorsements  from  the  time  of  their 
payment,  until  the  year  has  elapsed,  or  until  an  endorsement  is 
made  beyond  the  year,  which  with  the  preceding  endorsement  or 
^endorsements  and  its  or  their  intercut,  eqnah  or  exceeds  the  in- 
merest  due  at  snch  endorsement. 


27d        INTEREST  ON  BONDS  AND  OBLIGATIONS. 

4.  Interest  is  not  allowed  on  interest,  because  the  deduction, 
when  made,  is  intended  to  pay  the  interest  then  due,  and  the  ex- 
cess, if  any,  goes  to  reduce  the  principal. 

6.  The  design  of  the  rule  is  to  give  the  interest  as  nearly  annt.'- 
aVIy  as  the  endorsements  will  admit.  But  if  endorsements  are  not 
made  on  the  oblip^ation,  the  rule  implies  that  the  interest  is 
not  due  until  the  obligation  is  paid  ;  and,  it  is  well  known,  that  the 
Courts  will  not  sustain  an  action  for  the  payment  of  the  interest 
from  year  to  year,  uniess  the  obligation  contains  the  express  con- 
dition that  the  interest  shall  be  paid  annually. 

Note.  In  (he  "  Scholar's  Adthmetick,"  and  in  the  **  Mercan- 
tile Arithmetick,"  the  Rule,  said  io  be-established  by  the  Courts  of 
JMassachusetts,  is  precisely  the  same  as  that  established  in  the  State 
ef  New  York,  which  will  be  found  on  a  following  page,  It  will  b6 
evident  from  a  comparison  of  the  preceding  Rule  and  that  of  New 
York,  that  the  computation  of  interest  ougfi^  to  be  considerably  dif- 
ferent in  the  two  States. 

Rule  III. 

The  f<dlow»ng  Rule  was  established  by  the  Superior  Court  of 
Connecticut  in  1784. 

"  Com[)ute  the  interest  to  the  time  of  the  first  payment,  if 
that  be  one  year  or  more  from  the  time  the  interest  commenced,^ 
add  it  to  the.  principal,  and  deduct  the  payment  from  the  sum  total. 
If  there  be  after  payments  made,  compute  the  interest  on  the  bal- 
ance due  to  the  next  payment,  and  then  deduct  tbe  payment  as. 
stbove  ;  and,  in  like  manner  from  one  payment  to  another,  till  all  the 
payments  are  absorbed  ;  provided^  the  time  between  one  payment 
an<i  another  be  one  year  or  more.  But  if  any  payment  be  made  be- 
lore  one  year's  interest  hath  accrued,  then  compute  the  interest  on 
the  principal  sum  due  on  the  oblig;  tion  for  one  year,  add  it  to  the 
principal  ;  and  compute  the  interest  on  the  sum  paid,  from  the  time 
jt  was  paid,  up  to  the  end  of  Ihe  year,  add  it  to  the  sum  paid,  and  de- 
duct that  sum  from  tlie  principal  and  interest,  added  as  above.  If 
any  payments  be  made  of  a  less  sura  than  the  interest  arisen  at  the 
time  ot"  such  payment,  no  interest  is  to  be  computed  but  anly  on 
i\w.  principal  sum,  for  any  period." 

This  l^nle  involves  the  same  principles  as  Rule  II.  with  the  fol- 
lowing, vi',:. 

That  if  an  endorsement  be  made  after  a  year's  interest  has  ac- 
crued, but  which  i^  loss  than  this  interest,  it  shall  not  bear  interest 
Irlje  the  endorsements  made  before  a  year's  interest  ha,s  become 
flue. 

Rule   IV. 

The  f'^ilowing  Hole  is  established  for  the  practice  of  the  Courts 
j;>  the  State  of  New  York. 

"  The  Rule  for  ca-^ling  Interest,  when  partial  payments  have 
Itcvn  rnadp,  is  to  apply  the  payment,  in  the  fir.«t  place,  to  the  dis- 
charge of  the  interest  then  due.  If  the  payment  exceeds  the  in- 
terest, the  surplus  goes  towards  discharging  the  principal,  and  the 


INTEREST  ON  BONDS  AND  OBLIGATIONS.        271 

subsequent  interest  is  to  be  computed  on  the  balance  of  principal 
remaining  due.  If  the  payment  be  less  than  the  interest,  the  sur- 
plus of  interest  must  not  be  taken  to  augment  the  principal;  but 
interest  continues  on  the  former  principal  until  the  period  wbea 
the  payments,  taken  together,  exceed  the  interest  due,  and  then 
the  surplus  is  to  be  applied  towards  discharging  the  principal ;  and 
interest  is  to  be  computed  on  the  balance  of  principal  as  aforesaid." 
[Johnson's  Chancery  Reports,  Vol.  I.  page  17.] 
This  Rule  contains  the  following  principles. 

1.  rhat  Interest  is  due  at  any  time  when  a  payment  is  made,  if 
the  payment  is  equal  to  the  interest  to  that  time  :  if  not,  the  pay- 
ment is  to  be  added  to  the  following  payment  or  payments,  until 
their  sum  is  equal  to  or  exceeds  the  interest  then  accrued,  and  the 
balance  is  a  new  principal. 

2.  Interest  is  not  allowed  on  any  entlorsement. 

3.  Interest  is  not  taken  on  interest,  because  the  interest  is  due 
when  the  endorsement  or  endorsements  are  made. 

Note.     Example  18  in  Simple  Interest  is  calculated  by  this  rule. 

Example.  I. 
To  be  calculated  according  to  the  preceding  Rules. 
For  value  received  I  promise  to  pay  James  Penny  or  order  one 
thousand  dollars  on  demand  with  interest.  James  Gold. 

January  1st,  1815. 
The  endorsements  were, 
March  1,  1816,  received  on  the  above  Note,  75  dolls. 

July  1,  1816, 20     do. 

Sept.  1,  1817,      .         -         -         .         -         20     <lo. 

Nov.  1,  1817, 750     do. 

March  1,  1818, 100     do. 

What  is  the  balance  due  July  1st,  181B,  interest  being  allowed 
kt  6  per  cent. 

$  By  Rule  I. 

1000       Principal, 
210       Interest  to  July  1,  18J8,  being  3} years, 

1210       Amount  to  do. 


75       1st  endorsement  March  1,  1816, 

10-50  Interest  to  July  1,  1818,  being  2^  years-, 


85*50  Amount. 


20       2nd  endorsement  July  1,1816, 
2-40  Interest  to  July  1,  J818, 


22-40  Amount. 


20      3d  endorsement  Sept.  1,  1817, 
1       Interest  to  July  1,  1818,  being  XO  raontli^ . 


21       Amount. 


2n       INTEREST'  ON  BONDS  AND  OBLlGATlONSv 

750       4lh  endorsement  Nov.  1,  1817, 
30       Interest  to  July  1, 1813,  being  8  months. 


780       Amount. 


100       5th  endorsement  March  1,  1818, 
2       Interest  to  July  1,  1818,  being  4  months^ 


102       Amount. 
1010-90  Amount  of  the  sevieral  endorsements, 

199-10  Balance  due  July  1,  1818. 

$  By  Rule  II. 

1000       Principal, 

70       Interest  to  1st  endorsement  March  1,181G,  being  14 

[months. 

1070       Amount  to  do. 

75       Ist  endorsement — deduct. 


995       New  principal  March  1,  1816, 
99-50  Interest  to  4th  endorsement  Nov.  1,1817,  being  20mOc 


1094-50  Amount  to  do. 

20       2nd  endorsement  July  1,  1816, 
1-60  Interest  to  Nov.  1,  1817,  being  16  months, 


21-60  Amount  to  do. 


20       3d  indorsement  Sept  1,  1817, 

•20  Interest  to  Nov.  1,  1817,  being  2  monthSj 


20-20  Amount  to  do. 


'50       4th  endorsement  Nov.  1,  1817, 

41-80  Sumofthe  amounts  of  the  two  preceding  endorsements. 


791-80  To  be  subtracted  from  1094-50. 

302  70    Netv  principal  Nov.  .1,1817, 
12- 108  Interest  to  July  1,  1818,  being  8  months, 


,314-808  Amount  to  do. 


100         5th  endorsement  March  1,  1818, 

2         Interest  to  July  1,  1818,  being  4  months, 


102        Amount  to  do.  to  be  deducted  from  314-80S. 


212  808  Balance  due  July  1,  1818. 
19910         Do.  by  rule  1. 


13  708  Difference, 


INTEREST  ON  BONDS  AND  OBLIGATIONS. 

$  By  Rule  III. 

2000  Principal, 

70  Interest  to  1st  endorsement^ 


1070  Amount. 

75  1st  Endorsement, 


995  New  Principal  March  1,  1816, 

99-50     Interest  to  4th  endorsement,  being  20  month?, 


1094-50     Amount  to  do. 


20  2nd  endorsement, 

1-60     Interest  to  4th  endorsement, 


21-60     Amount  to  do. 

20  3d  endorsement,  which  does  not  bear  interest,  because 

one  year's  interest  has  accrued  on  new  principal. 
750  4th  endorsement, 


791-60     Deduct  from  1094-50, 


302-90     New  principal  Nov.  1,  1817, 
12-31-6  Interest  to  July  1,  1818,  being  8  monthSp 


315-21-6  Amount  to  do. 


100  5th  endorsement, 

2  Interest  to  July  1,  1818,  being  4  months. 


102  Amount— deduct  from  315-21-6. 


213-21  6  B-alance  due  July  1,  10 
212-80-8  Do.  by  Rules.  ' 


0-40  8  Difference, 
13- 11-6  Do.  from  result  of  Rule  I. 

i  By  Rule  IV. 

1000  Principal, 

70  Interest  to  1st  endorsement,  being  14  months, 

1070  Amount  to.  do. 

75  1st  endorsement. 


995  New  principal  March  1,  .1816, 

19-90     Interest  to  July;,  1816, 


1014-90     Amount  to  do. 

20  2nd  endorsement  July  1,  1816. 

994-90     New  Principal        do. 

L  1 


274        INTEREST  ON  BONDS  AND  OBLIGATION^! 
79-59-2  Interest  to  Nov.  1,  1C17» 


1074-49-2  Amount  to  do. 


20  3ti  endorsement, 

750  4th  do.  Nov.  1,  181 


770  Sum  of  these  endorsementt'. 


304-49-2  New  principal  Nov.  1,  181" 
6  08'9  interest  to  March  1,  1818, 


310  58-1  Amount  to  do. 

100  5lh  endorsement  do. 


210-681  New  principal  do. 

4-21-2  Interest  to  July  1,  18M, 


214-79  3  Balance  due  do. 


15-69  3  Difference  between  this  result  and  that  bv  Rule  L 

1  98-5  do.                                            and                Rule  11. 

1-77  6  do.                                          and               Rule  UU 

$  By  Rule  IV.  at  7  per  cent. 

1000  Principal, 

198-33i  Interest  to  4lh  endorsement  Nov.  1,  1817, being  34m 

1198-33i  Amount  to                                do. 

865  Sum  ofthe  first  four  endorsements. 


333-331      New  principal  Nov.  1,  1817, 
7771      Interest ta  March  1,  1818, 


34D11        Amount. 

100  6th  endorsement, 


241'11        New  principal, 

5  62  59  Interest  to  July  1,  18 1b\ 


246-73  59  Balance  due  do. 

Example  2. 
On  Jan.  1,  1816,  Samuel  Trusty  owed  me  776l.,  ou  wliich  I  war 
to  receive  interej^f  at  7  per  cent.  On  July  1,  he  paid  me  1001.; 
on  Jan.  1  1817,  251.;  -on  Sept.  1,251.;  on  March  1,  1818,251.'; 
on  July  1,  1819,  101.;  on  Sept.  1,  3901.;  and  on  Jan.  1,  182K 
the  balance  :  VVIjat  was  the  balance  by  the  four  preceding  rules 
and  uiiaX  is  the  whole  interest  paid  ? 

p     ..   ,     .    (  Balance  due  £384  5081. 

iiy  i\uic  i.  j  \Yj,^j^,  interest  paidi:f84-508f 

P     p    ,  (  Balance  £410  3444. 

J^y  ixuie  z,  <  ^^^^^jg  interest  paid£2103544- 


DISCOUxNT.  ^75 

Balance  £410  471 2.2|. 
terest  £2104718^.' 


By  Rale  3,  j  f^ 

V     D   I     A    S  Balance  £418-487+. 
By  Rule  4.  |  ,^,^^^^,  ^^^^  .  j2^_ 


Example  3. 
I  gave  a  promissory  note  to  A.  D.  March  1,  1817,  for  ^350. 
On  Dec.  1,  I  paid  gl30  ;  on  April  1,  18J8,  $35;  on  June  l,glO; 
Jan.  1,  1819,^40;  Dec.  1,^65  ;  and  March  1,  1820,  a  settlement 
was  made.  To  whom  was  a  balance  due,  and  how  much,  comput- 
ing interest  at  6  per  cent,  by  the  last  three  rules  ?     Ans. 


DISCOUNT 

IS  an  allowance  made  for  the  payment  of  any  sum  of  monc}*, 
;>erore  it  becomes  due,  and  is  the  difference  between  that  sum,  due 
some  iimo  hence,  and  its  present  worth. 

The  present  worth  of  any  sum  or  debt,  due  some  time  hence,  is 
such  a  sum,  as  if  put  to  interest,  would  in  that  time  an<i  at  the  rate 
|)er  cent,  for  which  the  discount  is  to  be  made,  amount  to  the  sum 
i)V  debt  then  due. 

Rule  I  * 
As  the  amount  of  £100  for  the  given  rate  and  time  is  to  £100  ; 
so  is  the  given  sum  or  debt  to  the  present  xvorth. 

*  That  an  allowance  ought  to  be  made  for  paying  money  before  it  becomes 
iue,  which  is  supposed  to  bear  no  interest  till  after  it  is  due,  is  very  reasonable  ; 
■  >\'  if  I  keep  the  moniey  in  my  own  hands  till  the  debt  shall  become  due,  it  is  plain 
i  .nay  make  an  advantage  of  it  by  putting  it  out  to  interest  for  that  time ;  but  if 
1  pay  it  before  it  is  due,  I  give  that  benefit  to  another ;  therefore  we  have  only 
lu  inquire  what  discount  ought  to  be  allov/ed.  And  here,  many  suppose  that, 
since  by  not  paying  till  it  becomes  d'ue  tlicy  may  employ  it  at  interest ;  there- 
fore, by  paying  it  before  due-,  the)^  sliall  lose  that  interest,  apd  for  that  reason  all 
such  interest  ought  to  be  discounted  ;  but  the  supposition  is  false,  for  they  caiv 
not  be  said  to  lose  that  interest  till  the  time  arrives,  when  the  debt  becomes  due ; 
whereas  we  are  to  consider  what  woul4  be  lost,  at  present,  by  paying  the  debt 
before  it  becomes  due  ;  this  can,  in  point  of  equity,  be  no  otlier  than  such  asuD% 
which  being  put  out  at  iviterest  till  the  debt  ^Iiall  become  due,  would  amount  to 
the  interest  of  the  debt  for  the  sarr.e  time.  It  is"  besides  plain,  that  the  advan- 
tage arising  from  discharging  a  debt  due  some  tim.e  hence,  by  a  present  payment, 
according  to  the  principles  above  mentioned,  is  exactly  the  same  as  employing 
the  whole  sum  at  interest  till  the  time  when  the  debt  becomes  due,  arrives  : 
for,  if  the  discount  allowed  for  present  payment  be  put  out  at  interest  ibr 
that  time,  its  amount  will  be  the  same  as  the  interest  of  the  vholc  debt  for  the 
same  time  ;  thus  the  discount  of  ij  106,  due  one  year  hence,  reckohihg  interest 
at  £G  per  cent,  will  be  £6,  and  £6  put  out  to  interest  at  £6  per  cent,  for  one 
year,  vrill  amount  to  £6  7s.  2id.  Vv^iich  is  exactly  equal  to  the  inierest  of  jG106» 
tor  one  year  at  £6  per  cent. 

The  truth  of  the  rule  for  working  is  eviden^t  from  the  nature  of  Simple  Inter- 
(^<  t ;  for  since  the  debt  may  be  considered  as  the  amount  of  some  principal  (call- 
.'  1  here  the  present  wf)rth)  at  a  certain  rate  per  cent,  and  for  the  given  time, 
■Jjat  amount  mii?t  be  iii  the  ?a:'^::  r-.-  ;>->ruoa  eitlior  io  it-  pjincipal  or  interest, 


S76  •  DISCOUiST. 

Subtract  the  present  worth  from  the  given  sum,  and. the  remain- 
der will  be  the  discount  required. 

Or,  As  the  amount  of  £100  for  the  given  rate  and  time,  is  to  the 
interest  of  £100  for  that  time  :  so  is  the  given  sum  or  debt  to  the 
discount  required. 

Or,  In  federal  money,  divide  the  given  sum  by  the  amount  of 
glOO  for  the  given  time  and  rate  ;  point  off  from  the  right  of  the 
quotient,  two  places  less  than  in  division  of  decimals  for  the  pren- 
eat  worth. 


i    ^xru-t :,  iU-.  j:>. ,.  ^e>  '^-'ooo   i  <s.  f  due  2  years  hence  at  5-i 


Examples. 

What  is  the  discount  of  \  ^^^^^^  11^'  \ 

^g2119  50c.  5     per  cent,  per  annum 

Interest  of  £100  per  annum=5  10 

2  years. 


n 

Add  100 

£         £      s.       £   ?.  d. 

As  £111  :  11  ::  635   17  :  63  0  2  disc.  Ans. 
£       £     .    £     s.       £     s.     d. 
Or,  As  111  :  100  ::  635  17  :  672   16  9^  present  worth. 
£      «.        £     s.    d.       £   g.   d. 
And  635  17--572   16  9i==63  0  2i  discount. 

IN    FEDERAL   MONEY. 

S  $  $       c-       $      c.  m. 

As  111  :     11  ::  2119  60  :  210  04  Oi=disc')unt.     Or, 

2119-6x100 
As  111  :  100  ::  2119  60  : jjj =  ^^1909  45c.  9irn.=^' 

present  worth;  and  2119  5~1909'4696~210  0406=discount  as 
before. 

2119-5 
Or, -YY-~= 19-094596;  and  1909-4596^prescnt  worth,  as  be- 
fore. 

2.  What  is  the  present  worth  of  f'SbO  payable  in  half  a  year, 
discounting  at  6  per  cent,  per  annum?  Ans.  ^339  80c.  6m. 

3.  What  is  the  present  \vorth  of  £Qbj  due  15  months  hence,  at 
i$  6  per  cent,  per  annum  ?  Ans.  £60  9s.  3^d. 

4.  What  is  the  disconnt  on  £97  10s.  dne  January  22,  this  being 
September  7th,  reckoning  interest  at  £6  oer  cent.  ? 

Ans.  £1    15   11. 

as  the  amount  of  any  other  sum,  at  the  same  rate,  and  for  the  same  time,  to  its' 
principal  or  interest. 

lu  coQimon  cases, the  interest  is  taken  for  the  discount,  the  parties  not  attend- 
ing to  the  real  difference  between  discount  and  interest.  Thus  ^'100  discounted 
in  this  way  fcr  a  year  at  6  per  cent,  or  ^6  is  taken  out,  and  the  person  receives 
.$94.  If  he  were  to  lend  the  $94  on  interest  lor  a  year  at  the  same  rate,  he 
/would  receive  pf  interest  $6  6^cts.  or  3B  cents  less  than  the  above  discount, 
which  is,  in  fact,  discounting  at  61 3  per  cent,  or  nearly  6'4  per  cent,  instead  c>i 
tj  pf;r  cent. 

lii  Biiiik  Discount,  the  interest  is  considered  as  the  discoiuit 


ABBREVIATIONS  IN  DISCOUNT.  277 

5.  What  ready  money  will  discharge  a  debt  of  ^1595  due  5 
ajonths  and  twenty  days  hence,  at  6  per  cent.  ? 

Ans.  gl541  32c.  6m. 

6.  Bought  a  quantity  ofgoods  for  §250,  ready  money,  and  sold 
Ihem  for  gSOO  payable  9  months  hence  :  What  was  the  gain,  in 
ready  money,  supposing  discount  to  be  made  at  6  per  cent.  ? 

Ans.  $37  8c.   l^m. 

7.  What  is  the  present  wojth  of  §960,  payable  as  follows  ;  viz. 
J  at  3  months,  ^  at  6  months,  and  the  rest  at  9  months,  supposing 
the  discount  to  be  made  at  6  per  cent.  ?  Ans.  g936  70c. 

Rule   H. 

As  any  sum  of  money,  at  6  per  cent,  per  annum,  will  double,  at 
simple  interest,  in  200  months  ;  therefore, 

To  200  add  the  number  of  months  for  which  the  discount  is 
required,  by  which  sum  divide  the  product  of  the  money  and  time, 
;  in  months,)  and  the  quotient  will  be  the  discount. 

Examples. 

h  What  is  the  discount  of  §150  75c.  for  a  year  ? 
200     150-75 
-f   12         Xl2 

§  c.  m. 

212)1809'Q0(8'633=:;:8-53  3  discount,  Ans. 
1696 


1130 

1060  150-75 

8-533 


700  

G36  Present  worth  142-217 


640 

636 


4 

2.  Wha.t  is  the  present  worth  of  §426  55c.  at  6  per  cent,  to  be 
paid  8  months  hence  ?  Ans.  §410  14c.  6m. 

3.  What  is  the  discount  of  £361   15s.  6d.  to  be  paid  1  year  and 
8  months  hence  ?  Ans.  J£32  17s,  9id. 


ABBREVUTIOm  IN  DISCOUNT. 

Any  principal  to  be  discounted  for  one  year,  at  any  of  the  fol- 
lowing rales,  (or  for  any  rate  and  time,  whose  product  is  equal  to 
any  of  the  following  rates)  being  multiplied  by  the  multiplier,  (if 
^ny,)  and  divided  by  the  corresponding  divisor,  the  quotient  will 
be  the  discount. 


2^7S. 


.ABBREVIATIONS  IN  DISCOUNT. 


Rates.* 

^-    11  ~-81  (or  by  9  and  9) 
2    -r-51 
21-^41 
'     4    -r-26 

5    -~21  (or  by  7  and  3) 
..a     u    -^63  andxS 
r  74-h43andx3 

-27  andx2  (orX2  and^9  and  3) 
-13 

-n 

-28  andx3  (orX3,  and~^7  and  4) 
-^  9 

Examples. 
1.  IJow  much  must  I  abate  of  £5394   jOs.  due  3  yeai^  hence^ 
a^t  2|  per  cent,  per  annum  ?  £5394  10s. 

■    ■       ■  2 


10 
U2i 


2-1 


X3  27~-3=9j  10789     0 

8,  therefore,  x2,  and-f'27  3)1198   lb  6i 


£399   11    10  Ans. 
2.  What  is  the  discount  of  $546  62c.  5m.  for  8i  years,  at  1  per 
cent,  per  annum,  (or  for  1  year,  at  8i  per  cent,  per  annum  ?) 

$     c.  m. 
13)546-62  5 


Ans.  g42-04  8 
3.  What  is  the  discount  of  ^125  at  IJ-  per  cent,  per  annum,  fo^ 
four  years,  (or,  at  4  per  cent,  per  annum,  for  1|  years  ?) 

Ans.  ^5  95c.  2m. 

*  These  contractions  are  obvious  from  any  example,  wrought  accordir.^  tc. 
the  General  Piule.     I'hus,  l6t  the  sum  to  be  discounted  be  300  dollar?. 
1.  At  1^  per  cent.    Then,  by  the  E.ule, 


lOU 

405 

'IT 


H  ::  $300  :  discount,  or, 
5 


405  :  5  ::  300  : = —  ,  and  is  the  Iluie 

405         81 
ol  :  J  ::  300  :  the  discount  req.uired. 

'2.  AtSpercoit.     Then, 
102  :  2  ::  300  :  discount,  or, 

300 
51  :  1  ::  300  :  discounts —  . 
51 

3.  At  2i  percent.     Then, 
t02'5  :  2-5  ::  300  :  discount,  or 

41  :.  1    ::  300  :  Ai:s\ver  required. 

4.  At  o  pe^r  cent..    Then, 
108-:  8  ::  300  :  discount,  or, 

27  ;  2  ;:  300  :  Answer  required. 
In  the  game  way,  may  all  the  contractions  be  made,     Tt.e  eonlractions  {uv' 
uvade  on  the  two  terms  of  tl^e  proportion  ^vhich  are  invariable,  when  the  rcls  is 
given. 


DISCOUNT  Bt  DECIMALS.  ^79 

DISCOUNT  BY  DECIMALS. 


* 


The  sum  io  he  discounted,  the  time  and  the  ratio  given,  to  find  tJie 
present  worth. 

Rule. 

Multiply  (he  ratio  by  the  time,  add  unity  to  the  product  for  a 
divisor ;  by  which  sum  divide  the  sum  to  be  discounted,  and  the 
jquotient  w^ill  be  tlie  present  worth.* 

Subtract  the  present  worth  from  the  principal,  or  sum  to  be  dis- 
counted, and  the  remainder  will  be  the  discount. 

Or,  as  the  amount  of  £l  for  the  given  lime,  is  to  £\,  so  is  llie 
"interest  of  the  debt  for  the  said  time,  to  tlie  discount  required. 

Subtract  the  discount  from  the  principal,  and  the  remainder  will 
be  the  present  worth. 

Examples. 
1.  What  is  the  present  worth  of  6001.  due  3  years  hence,  at  01. 
per  cent,  per  annum  ? 

First  iMethod. 
Ratio~06 
Multiply  by  the  time=    3 

rroduct=-18 
Add   1- 


Divisor=M8)6O0(508-4745  present  worth. 
600 
br, =:X508  9s.  5^d.  Ans. 

•06x3+1 
Present    worth=508*4745=JC50o  9s.    S^^d.  which,    subtracted 
from  the  principal,  will  give  the  discount— £91   lOs.  G^d. 

*  As  the  sum  to  be  discounted  is,  in  fact,  the  amount  of  some  principal  at  tiie 
S^iven  rate  and  time  ;  to  find  the  principal,  which  is  now  the  present  worthy  yovi 
have  only  to  employ  the  rule  for  Case  2,  in  Simple  hiterest  by  decimals.     This  is 

the  rule  in  the  text.    Thus  in  Ex.  1,  by  said  Case  2,  — j ; — ;-=the  -principal,  iii 

this  case  the  present  worth.     The  remninder  of  the  rule  is  evident  from  %vhat 
has  been  said  under  Discount,  Rule  1 .  •. 

jYo/e  1 .  In  the  method  used  in  Simple  Interest  by  Decimals,  yoU  may  easily 
find  rules  for  obtaining  either  of  the  four  terms,  ji^rcA'cnf  worthy  ratio,  time,  or  sjitn. 
io  be  discounted,  when  the  other  three  are  given. 

J^ote  2.  When  the  ratio  is  -06,  or  six  per  cent,  per  annum,  and  the  given  time 

yj,:i3  expressed  in  months,  if  the  debt  be  divided  by  unity  added  to  half  as  many 

c^  hundredths  of  an  unit  as  there  are  months  in  the  given  time,  the  quotient  will 

I;,  ^e  the  present  worth.     Thus  for  3  years  or  3G  month?,  the  divisor,  wo  have  jmr. 

Been  to  be  1+-06X3,— 1-f  •18,or  1  added  to  half  as  many  hundredths  as  there 

^re  months  ;  if  the  time  be  3^  years  or  42  months,  the  divisor  is  H--0Gx3"5.,-- 

1  +  -21  ;  if  10  months,  then  l-l--06XlA=l-f--05i  as  before ;  if  2  months,  thra 

l-h-C6  X-rj,=l-f -01 ;  if  1  month,  then  l-J-'OGX  i-*-,— 1-'-00>,  and    „^  c^ 


!gO  DISCOUNT  BY  DECIMALS. 

Second  Method. 

Ratio=-06 

Multiply  by     3     As  118  :  1  ::  108 
1 

Add  I-  M8)108'00(91*5254. 

Amount  of  11.  for  the 


given  time 

And  600X'06x3=108~interest  of  the  debt  for  the  given  time. — 
Discount=9I-5264=j£91  10s.  6d.  which  taken  from  the  principal 
will  leave  the  prestent  worth=£508  9s.  6d. 

2.  What  is  the  present  worth  of  §558  62c.  5m.  due  2  years 
hence,  at  4^  per  cent,  per  annum? 

First  Method, 

Ratio=-045 
X  Time=      2- 

•09 
-hi' 


Divisor=l'09)558'625(512'5— present  worth. 
545 


136 

109 


t)  1  o 
X.  1  o 

558625  545 

^)r,  — =g512*5Ans.  545 

•045x2+1  — - 

And  g558  625--g5 12-5=^46  12c.  5m.=discr)unt,  Or,  As  ^1-09 
(=amount  ofgl  for  the  given  time)  :  $1  ::  g50  27625  (=  interest 
of  the  debt  for  the  given  time)  :  g4G125=;li.'=coijnt  as  above. 
And,  g558-625—g46125=g512'5=present  worth,  as  above. 

3.  Required  the  present  worth  and  discount  of  §4125  50c.  at  6^ 
per  cent,  per  annum,  due  18  months  hence  ? 


.        C  present  worth  §3746   19c.  l^m. 
^°^-  I  discount  379  30      21. 


4.  What  ready  money  will  discharge  a  debt  of  13541.  8s.  due  "^ 
years,  3  months,  and  12  days  hence,  at  6^1.  per  cent,  per  annum  ? 

Ans.jeil35  7s.  9d. 


DUriES.  281 


DUTIES. 


DUTIES  are  assessed  upon  articles  imported  into  the  country, 
at  a  certain  rate  per  pound,  hundred,  ton,  gallon,  &c.  without  re- 
spect to  the  value  of  the  articles  ;  or  upon  articles  according  to 
their  actual  cost.  The  latter  are  called  ad  valorem  duties.  The 
duties  are  computed  in  the  former  case,  on  the  most  obvious  prin- 
ciples, as  will  be  seen  in  the  following 

Examples. 
1 .  Required  the  duty  on  987Ife  of  chocolate  at  3  cents  per  pound. 

3 


^29-61  cents,  Ans. 

2.  If  the  duty  on  molasses  is  5  cents  on  a  gallon  when  imported 
in  an  American  vessel,  and  10  per  cent,  more  in  a  foreign  vessel, 
what  is  the  duty  on  3960  gallons  in  both  vessels  ? 

Ans.  $197-50,  and  g217-25. 

3.  Required  the  duty  on  lOCwt.  3qrs.  14Ife  of  cordage  at  g2-25 
per  Cwt.  in  an  American  vessel,  and  at  10  per  cent,  more  in  a  for- 
eign vessel?  Ans.  g24-47  nearly,  and  §26-81. 

4.  What  is  the  duty  on  Ghhds.  of  brown  sugar,  weighing  53Cwt. 
2qrs.  201fe  tare  12lfe  per  100,  at  2|-cts.  per  pound  in  an  American 
vessel,  and  at  10  per  cent,  more  in  a  foreign  vessel? 

Ans.  §132,  and  gl45-20. 

Duties  ad  valorem  are  estimated  by  adding  20  per  cent,  to  the 
actual  cost  of  the  goods,  &c.  when  imported  from  or  beyond  the 
Cape  of  Good  Hope,  and  10  per  cent,  when  imported  trom  any 
other  places.  Insurance,  commission,  &c.  do  not  belong  to  the 
^actual  cost. 

The  duties  are  computed  in  the  following  obvious  manner. 
When  the  cost  is  reduced  to  Federal  Money,  add  the  per  cent,  to 
the  cost,  and  then  find  the  duty  per  cent,  ad  valorem. 

Examples. 
1.  What  will  be  the  duty  on  an  invoice  of  goods,  which  cost 
£786    15s.  sterling,  at  15  per  cent,  ad  valorem  when  imported  in 
an  American  vessel,  or  at  10  per  cent,  more  when  imported  in  a 
foreign  vessel  from  England  ? 

£     s.         $ 
786   15=:3493-17 
Add  10  per  ceot.=  349-317 

3842-487 


10  per  cent.=  384  2487 
6       do.       =^   192-12435 


15  per  ceut.=g576-37-305  Ans.  in  Am.  vessel 
16A  per  ct.=§G34-01  Ans.  in  Fgr.  ship. 

M  m 


282 


BARTER. 


2.  What  is  the  duty  on  goods,  which  cost  in  India,  g2780oO, 
imported  in  an  American  ship,  at  12 ^  per  cent,  ad  valorem? 

Ans.  $417-075. 


BARTER 

IS  the  exchanging  of  one   commodity  for  another,  and  teaclres 
traders  to  propdltion  their  quantities  vv'ithout  loss.* 

Case  i. 

When  the  quantity  of  one  commodity  is  given,  with  its  value,  or  that  of 
its  integer,  that  is,  of  lib.  lca?t.  lyd.  ^'C.  as  also  the  value  of  the 
integer  of  some  other  commodity ,  to  be  given  for  it,  to  find  the  quan-^ 
tity  of  ifiis ;  or,  having  the  quantity  thereof  given,  to  find  the  tat^ 
of  selling  it. 

Rule. 
Find  the  value  of  the  given  quantity  by  the  concisest  method, 
then  find  what  quantity  of  the  other,  at  the  rate  proposed,  you  may 
have  for  the  same  money :  Or,  if  the  quantity  be  given,  find  from 
thence  the  rate  of  selling  it.  Or,  As  the  quantity  of  one  article  is 
to  its  price,  so,  inversely,  is  the  quantity  of  the  other  to  its  price. 
Or,  as  the  price  of  one  article  is  to  its  quantity  ;  so  inversely fi»ihe 
price  of  the  other  to  its  quantity. 

Examples. 
1.  How  much  tea  at  9s.  6d.  per  fe  must  be  given  in  barter  for 
156  gallons  of  wirte,  at  12s.  3id.  per  gallon  ? 

Galls. 


3d. 


156 

12 


9s.  6d.  =  114d 


1872 
39 
6  6 

1917  G 
12 


23010 
d.  ft 


oz. 


As   114  :    1   ::  23010  ::   201    13//y  Ans. 
price,     quan.    price.       quan. 
s.    d.     gals.      s.  d.      ft      oz. 
Or,  As  12  3-^-  :   166  :,  9  G-  :  201   IS/^-V  Ans.  as  before. 


*  The  Rules  in  Barter  are  only  applications  of  the  Rule  of  Three,  and  are  ea- 
sily un(J.erstood. 


BARTER.  28J 

2.  How  much  clotb,  at  15s.  8d.  per  yard,  must  he  gjven  for  5cvyt. 
Sqrs.  191bs.  of  steel,  at  5  guineas  per  cwt?     Ans.  52yds.  3qrs.  2n. 

3.  Suppose  A  has  350  yards  of  linen,  at  25c.  per  yard,  which  he 
would  barter  with  B  for  sugar,  at  $5  per  cwt.  How  much  sugar 
will  the  linen  come  to  ?  Ans.  ITcwt.  2  qrs. 

4.  A  has  broadcloths  at  g44  per  piece,  and  B  has  mace,  at  gl 
42c.  per  ife :  How  many  pounds  of  piace  must  B  give  A  for  35 
pieces  of  cloth  ?  Ans.  1084|^ife. 

5.  A  has  7icwt.  of  sugar  at  12  cents  per  ife  for  which  B  gave 
Jiira  :i2^cwt.  of  flour  :     What  was  the  flour  rated  at  per  ife  ? 

4"^.  7c.  2m. 

CASE  H, 

If  the  quantites  of  two  commodities  be  given ^  and  the  rate  of  selling 
them,  to  find,  in  case  of  inequality,  how  much  of  some  other  com- 
modity mpst  be  given. 

Rule. 
Find  the  separate  values  of  the  two  comiHodjties  ;  subtract  the 
,ess  from  the  greater,  and  the  difference  will  be  the  amount  of 
^he  third  commodity,  whose  quality  and  rate  may  be  easily  found. 
t 

Examples. 

1.  Two  merchants  barter  ;  A  has  SOcwt.  of  cheese,  at  23s.  6d. 
per  cwt.  and  B  has  9  pieces  of  broadcloth,  at  31.  15s.  per  piece  : 
Which  must  receive  money  and  how  much? 

Ans.  B  must  pay  AjGl  10s. 

2.  A  and  B  would  barter  ;  A  has  150  bushels  of  wheat,  at  jjl 
25c.  per  bushel,  for  which  B  gives  G5  bushels  of  barley,  worth  62  ic. 
per  bushel,  and  the  balance  in  oats  at  37Jc.  per  bushel:  What 
quantity  of  oats  must  A  receive  from  B  ?        Ans.  391|  bushels. 

CASE  III. 

Sometimes,  in  bartering,  one  commodity  is  rated  above  the  ready  money 
price  ;  then,  to  find  the  bartering  price  of  the  other,  say, 

As  the  ready  money  price  of  the  one,  is  to  its  bartering  price  ;  so 
\i  that  of  the  other,  to  its  bartering  price  :  Next,  find  the  quanti- 
ty required,  according  to  either  the  bartering  or  ready  money  price. 

Examples. 

1.  A  has  ribbands  at  2s.  per  yard  ready  money  ;  but  in  barter 
ha  will  have  2s.  3d.  B  has  broadcloths  at  323.  6d.  per  yard  ready 
money  ;  at  what  rate  mu»t  B  value  his  cloth  per  yard,  to  be  equi- 
valent to  A's  bartering  price,  and  how  many  yards  of  ribband,  at  2s. 
3d.  per  yard,  must  then  be  given  by  A  for  488  yards  of  B's  broad- 
doth  ? 

Ans.  B's  broadcloth,  at£l    iGs.  6id.  per  yd.  7930  yds.  ribband. 

2.  A  and  B  barter  ;  A  has  150  gallons  of  brandy,  at  ^1  37|c.  per 
gallon  ready  money,  but  in  barter  he  will  have  $1  50c.  fB  has 


284  LOSS  AND  GAIN. 

linen  at  44c.  per  yard  ready  money  ;  how  mustB  sell  his  liuen  per 
yard  in  proportion  to  A's  bartering  price,  and  how  many  yards  are 
equal  to  A's  brandy  ? 

Ans.  barter  price  is  48c.  and  he  must  give  A  468yds.  3qrs. 

3.  F  and  Q  barter;  F  has  Irish  linen,  at  60c.  per  yard,  butiu^ 
barter  he  will  have  64c.  Q  delivers  him  broadcloth  at  $6  per 
yard,  worth  only  ^5  50c.  per  yard  :  Fray  which  has  the  advan- 
tage in  barter,  and  how  much  linen  does  P  give  Q,for  148  yards  of 
broadcloth  ? 

c.         c.  $    c.       $    c. 

As  60  :  64  ::  6  60  :  6  86|;  therefore  Q,  by  selling  at  $6  has 
the  advantage.     Then, 

$       yds.  c.  yds.  qrs. 

As  6  :   148  ::  64  :   1387  2  linen,  Ans. 

4.  A  has  2Q0  yards  of  linen,  at  Is.  6d.  ready  money  per  yard, 
which  he  barters  with  B,  at  Is.  9d.  per  yard,  taking  buttons  at  7id. 
per  gross,  which  are  worth  but  6d.  :  How  many  gross  of  buttons 
will  pay  for  the  linen,  who  gets  the  best  bargain,  and  by  how  much, 
both  in  the  whole,  and  per  cent.  ? 

Yd.    d.       Yds.         d.      .         d.  Gross,     d.     Gross.       Yd.   d.       Yds.    £ 

As  1  :  21  ::  200  :  4200.  As  71 :  1  ::  4200 :  660.  As  1  :  18  ::  200:  15. 

gr.  d.      gr.       £  [value  of  A's  linen. 

As  1  :  6  ::  660  :  14  value  of  B's  goods.  So  that  B  gains  ll.  of  A. 

£     £       £       £s-  d. 
As  14  :  1  ::  100  :  7  2  10  per  cent. 

5.  A  has  linen  clojh,  at  30c.  per  yard,  ready  money,  in  barter  36c. 
B  has  3610  yards  of  ribband,  at  22c.  per  yard  ready  money,  and 
would  have  of  A  ^200  in  ready  money,  and  the  rest  in  linen  cloth  ; 
what  rate  does  the  ribband  bear  in  barter  per  yard,  and  how  much 
linen  ntiust  A  give  B  ? 

Ans.  The  rate  of  ribband  is  26c-  4m.  per  yard,  and  B  naust  re- 
ceive 1980|  yards  of  linen,  and  ^200  in  cash. 


LOSS  AND  GAIN 

IS  an  excellent  rule,  by  which  merchants  and  traders  discover r 
their  profit,  or  loss  per  cent,  or  by  the  gross  :  It  also  instructs 
them  to  raise  or  fall  the  price  of  their  goods,  so  as  to  gain  or  lose 
go  much  percent,  &c.  The  rules  are  only  particular  applications 
of  the  Rule  of  Three. 

CASE  I. 

To  know  Zi)hat  is  gained  or  lost  per  cent. 

Rule. 
First  see  what  the  gain  or  lo^s  is,  by  subtraction  ;  then,  a§  the 
price  it  cost,  is  to  the  gain  or  loss  :  so  is  1001.  to  the  gain  or  loss 
T)nr  cent. 


LOSS  AND  GAIN. 


285 


Or,  in  federal  money,  annex  two  cyphers  to  the  gain  or  loss,  and 
^iivide  by  the  cost  for  the  gain  or  loss  per  cent. 

Examples. 
1.  If  I  buy  serge  at  90g.  per  yard,  and  sell  it  again  at  gl  2c.  per 
I   yard  :     What  do  I  gain  per  cent,  or  in  laying  out  glOO  ? 

c.      c.        S        J^ 
r  Sold  for  gl-02    As  9Q  :  12  ::  100  :  13^  per  cent,  gain,  Ans. 
^'    Cost  -90 


•Gain  -12  per  yard.  12-00 

Or,  1-02 — •90=12— gain  per  yard  ;  and =13i  per  cent,  gain, 

'^  [as  before. 

N.  B.  The  first  questions  in  the  several  cases,  serve  to  elucidate 
each  other. 

2.  If  I  buy  serge  at  $1  2c.  per  yard,  and  sell  it  again  at  90c. 
per  yard  :  What  do  I  lose  per  cent,  or  in  laying  out  JlOO? 

^  C.  ^    C.        C.  ^  ^      C.    DO. 

Cost         102     As  1  02  :  12  ::  100  :  11   76  5  per  cent,  loss,  Ans. 
Sold  for     -90 

—  12.00 


Loss 


12     Or,  y— 3"— 11-765  per  cent,  loss,  Ans.  as  before. 


3.  If  I  buy  a  cwt.  of  tobacco  for  £9  6s.  8d.  and  sell  it  again  at 
3s.  lOd.  per  ffe  do  1  gain  or  lose,  and  what  per  cent.  ? 

m  JS     s. ,  d. 

:    '  112  Sold  for  10     5     4 

-—  Cost  9.    6     8 

£  • 

j2d.|YV|l  I     4  value  at  2s.  per  ife.  0  18     8  gained  in  the  gross. 

0  18  8  value  at  2d.  per  Ife. 


10  5  4  value  at  Is.  lOd.  per  ife. 
£  s.  d.      s.    d.        £       £ 
As  9  6  8  :  18  8  ::  100  :  10     Ans.  10  per  cent.  gain. 
4.   A  draper  bought  60  yards  of  cloth  at  $4  50c.  per  yard,  an« 
38  yards  of  cloth  at  jJ2  50c.  per  yard,  and  sold  them,  one  with 
another,  at  <J4  25g.  per  yard  :   Did  he  gain  or  lose,  and  what  per 
cent.         60  yards  at      §4  50c.      per  yard     =     ^270 


38  yardii  at 


50         per  yard 


95 


98  yards  cost 365 

which  subtract  frona  98yds.  at  $4  25c.=;416-50 


gain  in  the  gross 


=  51-50 


$        $   c. 


$ 


Then,  as  365  :  51-50  ::  100  : 


.15000      $  c. 


365 


1411  gain  per  cent.  Ans. 


5.  Bought  sugar  at  6id.  per  ife  and  sold  it  atJ&2  3s.  9d.  per  cwt. 
What  was  the  gain  or  loss  per  cent.  ? 

fed.        ife     £  s.  d. 
As  1  :  6i  ::  112  :  3  0  ri 


^au 


LOSS  AND  GAIN. 


Prime  costX^ 
Sold  at  2 


8  per  cwt.      £>  s.  d. 

9  per  cwt.  as  3  0  8 


s.  d.         £      £    s.    d. 
16  11  ::  100  :  27   17  8^ 

[loss  per  cent.  Ans. 

Lost  .go  16  11  in  the  whole. 

6.  At  4s.  6d.  io  the  pound  profit :  How  much  per  cent.  ? 

X   ?.  a.       £      £   s. 
As  1  :  4  6  ::  100  :  22   10  Ans. 

7.  If  I  buy  candles  at  Is.  6d.  per  fe  and  sell  them  again  at  2s.  per 
fe  and  allow  3  months  for  payment :   What  do  1  gain  per  cent.  ? 

d.      d.  £        £      s.  d.  Mo.  j^     Mo.  £   s. 

As  18  :  24  ::  100  :  133  6  8  ;  then  by  discount.  As  12  :  6  ::  3  :  1  10 

£     s.      £    s.  £    s.  d.     £    s.     d. 

Then,  as  101   10  :  1    10  ::  133  6  8  :  1    19  4f,  which  taken  from 
X133  Gs.  8d.  Ieaves£l31  7s.  3id.  therefore,  Ans. £31   7s.  3|d. 

8.  If  I  buy  cloth  at  13s.  per  yard,  on  8  months  credit,  and  sell  it 
a^ain  at  12s.  ready  money,  do  1  gain,  or  lose,  and  what  per  cent.  ? 

Ans.  lost  £4  per  cent,  or  6d.  in  the  yard. 

9.  If  I  buy  gloves  at^l  25c.  per  pair:  How  long  credit  mu^t  1 
have,  to  gain  ^13  per  cent,  when  I  sell  them  at  $1  36c.  per  pair? 

'^   c.        ^  c.         c.         ^         ^  c 
Sold  at        VsG  As  1-25  :  -U  :  100  :  8-80  gain  per  cent.  rdy.  mo. 
Prime  cost  1-2^  $     ^  c.     g  c. 

Then,  13— 8-80=4'20  Now, 

Gained  -11  per  pair.  $    Mo.     g  c.  Mo.  days. 

As  6  :  12  ::  4-20  :  8   12  Ans.      - 
In  casting  up  tlie  a^i^'Jiit  of  goods  bought,  imported  or  export- 
ed :  to  the  prime  cost  of  such  goods  we  must  add  all  the  charges 
upon  them,  in  order  to  fix  the  price  they  stand  ns  in. 

10.  Suppose  1  import  from  France,  12  bales  of  cloth,  containing 
10  pieces  each,  which,  with  the  charges  there,  amounted  to  J360  : 
I  pay  duty  here  92c.  per  piece,  for  freight  §12  and  portage  gl 
25c.  ;  What  does  it  stand  me  in  per  piece,  and  how  must  1  sell  it 
per  piece  to  gain  §10  per  cent.  ? 

Ans.  §4  43  3  the  price  at  which  it  must  be  sold  per  piece, 

CASE  IL 

To  knoio  hojp  a  commodiiy  must  be  soldy  to  gain  or  lose  so  much  per' 

cent. 

KuLF.. 

As  £100  is  to  the  price  ;  so  is  £100  with  the  profit  added,  or  loss 
Subtracted,  lo  the  gaining  or  losing  price.     Or, 

In  federal  money,  multiply  100  dollars  added  to  the  gain,  or  less 
by  the  loss  per  cent,  by  the  cost ;  and  pointing  off  the  two  i  ight 
hand  figures  of  the  product  gives  the  answer. 

Examples. 

1.  If  I  buy  a  quantity  of  serge   at  90c.  per  yard:  How  R)ust  1 
It  it  per  yard  to  gain  13}  per  cent.  ? 

**  s'        C.  C.         *J(  c 

As  100  :  !!3  33}  ::  00  :  1  2  Ans. 


LOSS  AND  GAm.  287 

:  $      '*•        G.        $ 

.Or,  113  33ix90=102;  and  pointing  oflf  two  right  hadd  places, 
^1  02,  Ans.  as  before. 

2.  If  a  barrel  of  powder  cost  J£4,  how  must  it  be  sold  to  lose 
X 10  per  cent.  ?     £      £      £  Or  thus  : 

As  100  :  4  ::  90  90 

4  4 

100)360(3  £3160 

300  20 


60  9.12|00 

20 

100)1200(12  Ans.  £3   i2s. 
3.  Bought  cloth,  at  $2  50c.  per  yard,  which  not  proving  so  good 
as  I  expected,  I  am  content  to  lose  17}  per  cent,  by  it  :  How  must 
-I sell  it  per  yard?  Ans.  $2  6c.  2Jm. 

'    4.  If  120ife  of  steel  cost  £7,  how  must  I  sell  it  per  Ife  to  gain  £13; 
per  cent.  ?  Ans.  Is.  4d.  per  Ife. 

5.  A  gentleman  bought  10  tons  of  iron  for  £200,  the  freight  and 
duties  came  to  £25,  and  his  own  charges  to  £8  6s.  8d.  ;  How  must 
he  sell  it  per  Ife  to  gain  £20  per  cent,  by  it  ? 

£        £        £     s.  d.     £    s.    d.  £      s.  d.     £    s.   d.      £ 

As  100  :  20  ::  233  6  8  :  46   13  4  Then,  233  6  8-f46   13  4=280, 
Tons.     £         lb.     d. 
As  10  :  280  ::  1  :  3  per  ft  Ans. 

6.  If  a  bag  of  cotton,  weighing  8cwt.  Oqrs.  201fe  cost  §45  55c. 
how  must  it  be  sold  per  cwt.  to  lose  g8  per  cent.  ? 

/-  Ans.  $5  12c.  3m. 

7.  Bought  fish  in  Newburyport,  at  10s.  per  quintal,  and  sold  it 
at  Philadelphia,  at  17s.  6d.  per  quintal  ;  now,  allowing  the  charges 
at  an  average,  or  one  with  another,  to  be  2s.  6d.  per  quintal,  and 
considering  I  must  lose  £20  per  cent,  by  remitting  my  money 
home  ;  what  do  I  gain  per  cent.  ? 

Selling  price  17  6  Philadelphia  currenc}^  per  quintal. 
Charges     2  6  ditto. 


15  0  ditto. 
£       5.        £      ?. 
As  100  :  15  ::  80  :  12  New  England  currency. 
Sold  at  12s.  per  quintal. 

Bought  at   10s.  per  quintal. 

Gained  2s.  per  quintal, 

s.      s.        £        £ 
As  10  :  2  ::  100  :  20  per  cent,  gained,  Ans. 
0.  Bought  50  gallons  of  brandy,  at  75c.  per  gallon,  but,  by  acci- 
dent, 10  gallons  leaked  out:  At  what  rate  must  I  sell  the  remain- 
der per  gallon,  to  gain  upon  the  whole  prime  cost,  at  the  rate  o; 
10  per  cent.  ?  Ans,  $1  3c.  l^m. 


288 


LOSS  AND  GAIN. 


CASE  III. 

When  there  is  gain  or  loss  per  cent,  to  know  what  the  commodity  cost- 

Rule. 

As  .£100  with  the  gain  per  cent,  added,  or  loss  per  cent,  sub- 
'racted,  is  to  the  price  ;  so  is  £100  to  the  prime  cost.     Or, 

In  federal  money,  divide  the  price  with  two  cyphers  annexed  by 
§100  added  to  the  gain,  or  less  by  the  loss,  per  cent,  for  the  answer. 

Examples. 

1.  in  yard  of  cloth  be  sold,  at  ^1  2c.  and  there  is  gained  13} 
per  cent,  what  did  the  yard  cost  ? 

*  4    C.  ^  C 

As  T0O+T3I  :  1  2  ::  100  :  90  prime  cost,  Ans. 

10200 
Or,  ;j^^^_^='9,  Ans.  as  before. 

2.  If  12  yards  of  cloth  are  sold  at  15s.  per  yard,  and  there  vi 
£1  10s.  loss  per  cent  in  the  sale  :  What  is  the  prime  cost  of  the 
whole. 

Yds.    s.      Yds.   £  £    s.      £        £      £    s.   u. 

As  1  :  15  ::  12  :  9     As  92   10  :  9  ::  100  :  9   14  7  Ans. 

3.  If  401fe  of  chocolate  be  sold  at  25c.  per  Ife  and  I  gain  9  per 
cent,  what  did  the  whole  cost  me  ?  Ans.  $9  17c.  4m. -p 

4.  If  19icwt.  sugar  be  sold  at  gl4  50c.  per  cwt.  and  I  gain  gl5 
per  cent. :  What  did  it  cost  per  cwt.  ?  Ans.  gl2  60c.  8m. 

CASE  IV. 

If  by  'wares  sold  at  such  a  rate,  there  is  so  much  gcined  or  lost  per 
cent,  to  know  what  would  be  gained  or  lost  per  cent,  if  sold  at 
another  rate. 


R 


ULE. 


As  the  first  price  is  to  £100  with  the  profit  per  cent,  added,  or 
loss  per  cent,  subtracted  ;  so  is  the  other  price  to  the  gain  or  loss 
per  cent,  at  the  other  rate. 

N.  B.  If  your  answer  exceed  100,  the  excess  is  your  gain  per 
cent,  but  if  it  be  less  than  100,  the  deficiency  is  your  loss  per  cent. 

Examples. 

1.  If  cloth,  sold  at  ^1  2c.  per  yard,  be  13J-  profit  per  cent^ 
what  gain  or  loss  per  cent,  shall  I  have,  if  I  sell  the  same  at  ?0c. 
per  yard  ? 

$  c.        $  c.         $ 

As  1   2  :  113i  ::  90  :  ,100 
And,  100 — 100=0,  Ans.  I  neither  gain,  nor  lose. 

2.  If  cloth,  sold  at  4s.  per  yard,  be  £10  per  cent,  profit :  What 
shall  I  gain  or  tose  per  cent,  if  sold  at  3s.  6d.  per  yard? 


EQUATION  OF  PAYMENTS,  289 


3, 

£       s.  d. 

As  4: 

110  ::  3  6 

12 

12 

JS        £       £ 

,^ 

— 

Then,  100— 961=3^ 

48 

42 

no 

\ 


48)4620(961  Ans.  I  lostJCSf  per  cent,  by  (he  last  sale. 
432 

300 

288 

12 

3.  If  I  sell  a  gallon  of  wine  for  ^I  50c.  and  thereby  lose  12  per 
cent.  :  What  shall  I  gain  or  lose  per  cent,  if  I  sell  4  gallons  of  the 
^anie  wine  for  ^6  75c.  ?  Ans.  1  per  cent.  loss. 

4.  1  sold  a  watch  for  501.  and  by  so  doing,  lost  171.  per  cent- 
whereas  in  trading  I  ought  to  have  cleared  201.  per  cent.  How 
much  was  it  sold  under  its  real  value  ?  Ans.  221.  5s.  9|d. 


EQUATION  OF  PAYMENTS 

IS  the  finding  a  time  to  pay,  at  once,  several  debts,  due  at  differ- 
ent tiroes,  so  that  no  loss  shall  be  sustained  by  either  party. 

Rule  1.* 
Multiply  each  payment  by  the  time  at  which  it  is  due  ;  then  di- 
vide the  sum  of  the  products  by  the  sum  of  the  payments,  and  the 
quotient  will  be  the  equaled  time,  or  that  required. 

*  This  rule  is  founded  upon  H  supposition  that  the  sum  of  the  interests  of  the 
several  debts,  which  are  payable  before  the  equated  time,  from  their  terms  to 
that  time,  ought  to  be  equal  to  the  sum  of  the  interests  of  Ihe  debts  payable  after 
the  e  luated  time,  from  that  time  to  their  terms.  Some,  who  defend  this  princi- 
ple, have  endeavoured  to  prove  it  to  be  ri2;ht  by  this  argument ;  that  what  is 
{gained  by  keeping  some  of  the  debts  after  they  are  due,  is  lost  by  paying  others 
before  they  are  due  ;  but  this  cannot  be  the  case  ;  for  though,  by  keeping  a  debt 
after  it  is  due,  tiiere  is  gained  the  interest  of  it  for  that  time  ;  yet,  by  paying  a. 
debt  before  it  is  duo,  the  payer  does  not  lose  the  interest  for  that  time,  but  the 
discount  only,  which  is  less  tiian  the  interest,  ond  therefore  the  rule  is  not  accu- 
rately true  ;  however,  in  most  questions,  which  occur  in  business,  the  eiTour  is /o 
trilling,  that  it  will  always  be  made  use  of  as  the  most  elligible  method. 

From  the  principle  assumed  in  tl-iis  rule,  the  rule  may  be  derived  in  the  follow- 
ing manner.  Thus  in  Example  1,  where  W  months  is  found  to  be  the  equated  time, 
let  the  interest  be  supposed  at  any  rate,  as  6  per  cent.     Then  the  first  payment 

.'      ,         .            \.                                                                                100X0X8*-^ 
IS  to  be  at  mterest  for  8—6  or  2  mouthy  and  by  the  rule  for  interest, -— - 

=it.s  interest.     The  second  sum  i^  to  bo  on  interest  ^- — "■  of  1  nionth,  end 
N   n 


290  EQUATIOiN"  OF  PAYMENTS. 

Examples. 

1.  A  owes  B  g380  to  be  paid  as  follows,  viz.  glOO  in  6  monlh?, 

^120  iQ  7  months,  and  gl60  in   IQ  monihs  :     What  is  the  equatet^ 

time  for  the  payment  of  the  whole  debt  ? 

HK)x  C=  600 

120X  7=   840 

1(30X10=1600 


100+1204-160—380)3040(8  Months,  Aus. 
3040 

2.  A  owes  B  f04l.  15s.  to  be  paid  in  4 J  months,  1611.  to  be  paid 
in  3^  months,  and  1621.  6s.  to  be  paid  in  5  months  :  What  is  the 
equated  time  for  the  paj^ment  of  the  whole  ? 

Ans.  4  months  and  8  days. 

3.  There  rs  owing  to  a  merchant  9981.  to  be  paid,  1781.  ready 
money,  200K  at  3  months,  and  3201.  rn  8  months ;  I  demand  the  irr 
different  time  for  tbe  payment  of  the  vvhoFe  ?  Ans.  4^  months 

4.  The  sum  of  ^164  I6c.  6m.  is  to  be  paid,  ^in  6  months,  i  in  C 
months,  and  ■}  in  12  months  :  what  is  the  mean  time  for  the  pay- 
ment of  the  whole  ?  Ans.  7|  months. 

Rule  U. 
See,  by  rule  1st,  at  what  time  the  first  man,  mentioned,  ought  to 
pay  in  his  whole  money;  then,  as  his  money  is  to  his  time,  so  is 
the  other'^s  money,  to  bis  time,  inversely,  which,  when  found,  must 
be  added  to,  or  subtracted  from,  the  time  at  which  the  second  ought 
to  have  paid  in  his  money,  as  the  case  may  require,  and  the  sum? 
or  remainder,  will  be  the  true  time  of  the  second's  payment. 

Examples. 
1 .  P  is  indebted  to  Q,  $  t50  to  be  paid,  $50  at  4  monihs,  and  g  100 
at  8  months  :  Q.  owes  F  ^250  to  be  paid  at  10  months  :   tt  is  agreed 


=;it3  interest  to  the  equated  time.     The  sum'  of  the  interest  of 

100  ^ 

these  two  payineuts  is,  by  the  assumed  principle,  to  be  equal  to  the  interest  of 

.    160x6xiu — i:. 
the  third  payment,  or  j£lSO  for  10 — 8  or  2  months,  which  is — 

IOOxGxh"-^  ,   l^OxGxiT-^     160X6X10— H      ^^ 

Then, ' = ^ -—'    Now,  as  the  rate  per 

100  100  100  ^ 


cent,  and  100  will  be  factors  common  to  every  term  in  every  case,  they  may  be 
expunged  from  every  term,  and  then  we  have, 

lOOXH— 0  +  120  X;{-^=160X  10— B.  From  thio  equivalent  expression,  it  is 
easy  to  find  the  equated  time;  for,  100x8— 100x6-fl20Xt!— 120x7=100 
XlO— 160xr„    or   100x8-1-^-^0X8+160x3— 100x6-|-120x7-fl60xl0,    or, 

8  Xl(X)+lt'0+ 160—100x6+120x7+160x10,  and 

8=100X6+120X7+160X10      ,.,.,,,,        ^^ 

,  which  IS  the  rule.     The  same  may  be  shown 

loo+ix-o+ioo  ^ 

in  every  similar  case,  and  tlio  general  rule  inferred. 

This  rule  is  manifestly  incorrect.  The  iriie  rule  will  bo  given  in  Equation  o'.' 
I^aymenta  by  Decimal'-. 


Kt^UATION  OF  PAYMENTS  BY  PECIMALS.         291 

between  them  that  P  shall  make  preser*!  pay  of  his  whole  defct, 
and  that  Q,  shall  pay  his  so  much  the  sooner,  as  to  balance  that  fa- 
vour ;  I  demand  the  time  at  which  Qmust  pay  the  $250  reckonin,^ 
simple  interest. 
■  50X4=200 

100X8=800 

60-f-100=15lO)100|0(6|  months,  P*s  equated  time. 
90 

10 

L).     mo.      D.    mo.  mo.  mo.  mo. 

As  150  :  6|  ::  250  :  4.     Then,  10—4=6  time  ofQ's  payment. 

2.  A  merchant  has  1201.  due  to  him,  to  be  paid  at  7  months  ;  but 
the  debtor  agrees  to  pay  ^  ready  money,  and  ^at  4  months  ;  I  de- 
inand  the  time  he  must  have  to  pay  in  the  rest,  at  simple  interest^ 
§0  that  neither  party  may  have  the  advantage  of  the  other  ? 
Debt  £120 


1=60  must  be  paid  down. 
^=40  must  be  paid  at  4  months. 
i=20  unpaid. 
Now,  as  he  pays  601.  7  months,  and  401.  3  months  before  they  are 
respectively  due,  say,  as  the  interest  of  201.  for  1  month,  is  to    1 
month,  so  is  the  sum  of  the  interest  of  601.  for  7  mqnths,  and  of 
401.  for  3  months,  to  a  fourth  number,  which,  added  to  the  7  months, 
will  give  the  time  for  which  the  201.  ought  to  be  retained. 

Ans.  2  years  and  IQ  mouths. 

3.  A  merchant  has  §1200  due  to  him,  to  be  paid  i  at  2  months, 

}  at  3  months,  and  the  rest  at  6  months  ;  but  the  debtor  agrees  to 

pay  i  down  :    How  long  may  the  debtor  detain  the  other  half,  so 

that  neither  party  may  sustain  loss  ?        ' 

Now  as  i  was  paid  4^  months  before  it  was  due,  it  is  reasonable 
Jhat  he  should  detain  the  other  ^,  4^.  months  after  it  became  due, 
which  added,  gives  8|  months,  the  true  time  for  the  second  pay- 
ment. Equated  time=4^  months. 

EQUATIOJV  OF  PAYmNTS  BY  DECIMALS. 

Rule,* 
1.  To  the  sum  of  both  payments  add  the  continual  product  of 
the  tirst  payment,  the  ratio,  and  the  time  between  the  payments, 
and  call  this  the  first  number. 

*  Suppose  a  sum  of  money  be  due  immediately,  and  another  at  the  expiration 
of  a  certain  given  time  for^^arJ,  and  it  is  proposed  to  find  a  time,  so  that"  neither 
party  ghall  sustain  loss. 

Now,  it  is  plain  that  the  equated  time  must  fall  between  the  two  payments ; 
and  that  what  is  gotten  by  keeping  the  first  debt  after  it  is  due,  should  be  equal 
•J)  what  it  l(^st  by  paying  the  seccud  deU  before  it  is  dwe  ;  but  the  gain  arising 


£92         EQUATION'  OF  PAYMENTS  BY  DECIMALS. 

2.  Multiply  twice  the  first  payment  by  the  ratio,  and  call  this  the 
second  number. 

3.  Divide  the  first  number  by  the  second,  and  call  the  quotient 
the  third  number. 

4.  Call  the  square  of  the  third  number  the  fourth  number. 

5.  Divide  the  product  of  the  second  payment  and  time  between 
the  payments  by  the  product  of  the  first  payment  and  the  ratio,  and 
call  the  quotient  the  fit^th  number. 

6.  From  the  fourth  number  take  the  fifth,  and  call  the  square 
root  of  the  difference  the  sixth  number. 

7.  Then  the  difference  of  the  third  and  sixth  numbers  is  the 
equated  time,  after  the  first  payment. 

Examples. 
There  are  glOO  payable  in  2  years,  and  gl06  at  6  years  hence  ; 
what  is  the  equated  time,  allowing  simple  interest,  at  6  per  cent. 
per  annum  ? 

1st  payment=^100  1st  payment  100 

Ratio=-06  Multiply  by       2 

6-00  200 

Time  between  the  payments=4ys.  MnU.bytheratio=  06 


Add  both  payments=  j 
Div.  by   the  2d  num.  =  12)230=lst  number. 


24  12-00?=2dnuai- 

100 
106 


19-166-f-=3d  number. 
19-166+ 


3d  number  squared— 367-333556=;lth  number. 
2^1  payment-- 106 
Multiplied  by  the  time=^     4 

Ui  payment  >nult.  by  .he  raUo=6)42^.  \  ^'i^J^^r;::^^ 

70  666-f— 5th  number. 
From  the  4th  number=367  333556 
Take  the  6th  number=  70  666666 


296  668890(17-221sqr.rool~6thniifiK 
Carried  up. 

from  the  keeping  ofu  sum  of  money  after  it  is  due,  is  evidently  equal  to  tlie  in- 
terest of  the  debt  for  that  time  :  And  the  loss,  which  is  sustained  by  the  payiui 
of  a  sum  of  money  before  it  is  due,  is  evidently  equal  to  the  discount  of  the  del 
for  that  time  :  Therefore  it  is  obvious  that  the  debtor  must  retain  the  sum  in 
mediately  due,  or  the  first  payment,  till  its  interest  shall  be  equal  to  the  discouv.t 
of  the  second  sum  for  the  time  it  is  paid  before  due  ;  because  in  that  case  the  gaiii 
and  loss  will  be  equal,  and  consequently  neither  party  can  be  a  loser. 


EXCHANGE.  293 


From  the  3d  number=191 66        Brought  up. 
Take  the  Gth  number=17-224 


1  ■942==equated  time  from  the  first  pay- 
ment ;  therefore  3  942  years 
=3y.  11m.  9d.= whole  equat- 
ed time. 


100+ 106+ 100  X -06X4       100+ 106+ 100  X '06X4 

Or, 


106X412 

=1-942. 


100  X -06 1 


100X2X-06  100X2X-06 

2.  There  are  glOO  payable  one  year  hence,  and  gl06  to  be  paid 
six  years  hence  ;  what  is  the  equated  time,  computing  interest  at 
6  per  cent.  ?  Ans. 

3.  A  debt  of  ^1000  is  to  be  paid,  one  half  in  three  years  and 
the  other  half  in  6  years  ;"what  is  the  equated  time  for  paying 
both,  computing  interest  at  7  per  cent  ?  Ans. 


EXCHANGE. 


THE  object  of  Exchange  is  to  ascertain  what  sum  of  money 
ought  to  be  paid  in  one  country  for  a  sum  of  different  denomina- 
tions or  of  different  relative  value  received  in  another,  according 
to  the  course  of  exchange. 

The  par  of  exchange  respects  the  intrinsic  value  of  the  money 
of  different  countries  compared  with  each  other.  Thus  a  pound 
sterling  is  equal  to  4  dolls,  and  44  cents  in  the  United  States ;  the 
mark  banco  of  Hamburgh,  to  331  cents  ;  40  marks  banco  to  £3 
sterling.  If  the  exchange  be  made  at  the  intrinsic  value  of  the 
money  of  different  countries,  it  is  said  to  be  at  par ;  but  if  the  mo- 
ney of  one  country  be  estimated  at  less  or  more  than  its  intrinsic 
value,  the  exchange  is  said  to  be  above  par,  or  below  par. '^ 

Owing  to  o<ianges  in  the  course  of  trade,  to  demand  for  money, 
to  variations  in  the  relative  value  of  gold  and  silver,  &c.  the  rela- 
tive value  of  the  money  of  two  countries  is  liable  to  frequent  chan- 
ges. Hence  the  course  of  exchange^  or  the  current  price  of  ex- 
change, must  vary  with  these  circumstances,  and  be  sometimes 
above,  and  sometimes  below,  par.  Tables  of  the  course  of  ex- 
change are  published  daily  in  the  great  commercial  cities. 


*  The  Rules  under  Reduction  of  Coins  are  founded  on  the  par  of  exchange* 
For  the  reduction  of  the  Money,  and  Measures  of  most  commercial  countries  to 
Federal  and  Sterling  Money,  and  American  Measures,  see  also  the  Tables  of 
Money,  Length,  Capacity  and  Weight. 


m  EXCHANGE. 

1.  OF  GREAT  BRITAIN.* 
The  denominations  are  pounds,  shillings,  and  pence. 

Examples, 

1.  What  is  the  amount  in  Federal  Money  of  a  Bill  of  Exchange 
on  a  merchant  at  Liverpool  of  £133  sterling,  sold  in  New  York  at 
^  per  cent,  adrance  ?  £        ^    c. 

133=:59M1| 

295|f==^  per  cent. 

Amount  g5940G|  Ans. 

2.  In  Aug.  1821,  Bills  on  London,  bore  at  Boston  a  premium  of  8^ 
per  cent. ;  what  is  the  amount  of  a  bill  of  exchange  of  £250,  at  this 
rate,  in  Federal  Money,  and  what  is  the  value  of  a  pound  sterling 
at  this  course  of  exchange  ?  Aos.  Amount  gl205  65fcts. 

Value  of  a  pound  sterling  $4  82|cts. 

3.  A  Bill  of  Exchange  on  London  of  £90  sterling,  was  sold  at 
New  York,  at  36  shillings  New  York  currency  per  pound  sterling  ; 
what  was  its  amount  in  the  currency  of  New  York,  and  how  much 
above  or  Maw  pari 

£     ?.       £       £ 
As  1  :  36  ::  90  :  162  N.  Y.  currency. 
Now  £9  sterlings  £16  N.  Y.  currency,  or  20s.  sterling~35|  N.  Y» 
currency.     But  36— 33|=|^s.  N.  Y.  currency,  the  gain  on  every 
pound  sterling,  or £2  N.  Y.  in  the  whole. 

Then.,  as  35f8.  :  |  ::  100  :  1}  per  cent,  above  par. 
Or,        162—2  :  2  :;  100  :  lA  do. 

4.  The  invoice  of  gootls,  amounting  to  £170  10s.  sterling,  is 
sold  at  New  York  at  26  per  cent,  advance  ;t  what  is  the  amount 
in  Federal  Money  ?  Ans. 

*  The  Rules  on  which  Uye  operations  of  Exchang;e  are  performed,  are  obvi 
fu-5  from  the  rules  for  Reduction  of  Coins,  and  the  Rule  of  Three. 

t  To  reduce  sterliog  money  to  the  currency  of  New  England,  when  there  ^ 
&  e&rtain  per  cent,  advance,  merchants  use  the  following  method. 
For  12i  per  cent,  advance,  Inultiply  the  sterling  by  li 
50  ----.-.         la 

f5 .    .  li 

3U       :.----.       ^       i| 

bO -         '         -^  2 

C>2i         -------  2^ 

75     -         -         - 2i 

U7i 2i 

100    -        -        ~ 2^ 

125  -        - 3 

150     --------  34 

175 2|. 

200     -         - 4 

i  liese  multipliers  are  thus  formed.   Let  the  advance  be  25  per  cent,  on  £100  ; 

then,  as  25:=i  of  a  hundred,  lOOX  —  = —  =:tho  sum  with  the  advance.     This 

4         4 
IS  to  be  reduced  to  New  England  currency  by  increasing  it  by  one  third  of  itself. 

Thus  —-  X:^ — :="T<:r^^^^'3'  pounds ;  which  is  evidently  the  same  as  to  mu^- 
4       3X4       12 


tiply  100  by  1  J.    In  the  same  way  may  the  other  multipliers  be  four 


EXCHANGE.  ^95 

5.  A  Bill  of  Exchange  of  £75  16s.  is  sold  at  Boston  at  26s.  New 
England  currency  per  pound  sterling ;  ivhat  is  the  value  in  Fed 
nral  Money  of  a  pound  sterling  at  this  rate  of  exchange  ?     Ans. 

2,  OF  FRANCE. 

The  money  of  account  is  livres,  sols,  and  deniers. 
12  deniers  make     1  sol  or  shilling. 
20  sols  1  livre  or  pound. 

The  livre  is  estimated  at  18^-  cents  in  the  U.  S. 
The  crown  of  exchange  is  3  livres,  or  livres  tbumois,  aiid  i^ 
equal  to  55|  cents. 

The  present  money  of  account  is  francs  and  centimes  or  hun- 
dredths. 80  francs~8i  livres,  or  a  franc=||^  livre. 

1.  To  reduce  francs  to  livres,  or  the  contrary,  multiply  the  francs 
by  81  and  divide  the  product  by  80  for  livres  ;  or  Hvultiply  the  li- 
vres by  80  and  divide  the  product  by  81  for  francs. 

2156x81 
Thus  215G  francs= ^7; =2183  \Wtes  19  sols.     And  2341 

2341x80 
iivre9=    '   ^. =2312  francs,  09||  centimes. 

2.  To  reduce  livres  to  dollars  and  cents  :  multiply  the  livres  by 
the  <;ents  in  a  livre  at  the  course  of  exchange.  • 

Examples. 

1.  If  the  livre  be  20  cents  in  exchange,  what  is  the  amount  of 
2150  livres  in  Federal  money,  and  what  is  the  per  cent,  above  par 
at  this  exchange  ? 

Ans.  Amount  is  g430.     And  above  par  83*^  per  cent. 

2.  If  the  livre  be  18  cents  in  exchange,  required  the  amount  of 
3580  livres  16  sols,  in  dolls,  and  cents,  and  the  rate  per  cent,  be- 
low par. 

Ans.  644-54/^  cents,  and  2?^  per  cent.  belo\v  par. 

3.  If  a  crown  be  valued  in  exchange  at  18d.  sterling,  required 
the  livres  in  £100  sterling,  and  the  amount  also  iu  Federal  money 
at  par.  d.    liv.      £       liv. 

As  31ivres=l  crown,  18  :  3  ::  100  :  4000  and  4000x1 8-Ji=g740, 

4.  In  2583  francs,  how  many  dollars  ? 

"  2583x55A  =  1433dol]s.  561  cents. 

5.  A  bill  of  exchange  on  a  merchant  in  New  York  of  ^730  65cts= 
was  bought  at  Paris  at  I^  per  cent,  advance  ;  what  is  the  amount 
in  francs,  and  what  was  the  estimated  value  of  a  franc  at  this  es.^ 
change  ?  Ans. 

3.  OF  SPAIN. 
4  Maravadies  make        1  quarto. 
81  quartos=34  raarav,  1  rial  plate. 
8  rials  plate  1  piastre  or  current  dollar^ 

375  maravadies  1  ducat  of  exchange. 

Hard  or  plate  dollars  arc  88-/^  per  cent,  above  current  dollars  qt 
money  of  vellon,  o'™ 


^96  ElTHANGE. 

100  rials  pla(e     =188/^  rials  vellon. 
17     do.  =::32  do. 

The  rial  plate  is  10  cents,  and  the  rial  vellon  6  cents  in  the  U. 
States. 

To  reduce  rials  plate  to  rials  vellon,  or  the  contrary,  multiply 
the  rials  plate  by  32  and  divide  the  product  by  17,  for  rials  vellon  ; 
ov  multiply  the  rials  vellon  by  17  and  divide  the  product  by  32,  for 
rials  plate. 

1100x32 

1.  Thus  1100  rials  plate= jp^ rials  vellon=::2070i^  Ans. 

100x17 
And  100  rials  vellon= — ^^  rials  plate=53|  rials  plate.  Ani-'. 

Note.  The  rules  to  reduce  rials  plate  or  vellon  to  Federal  Mo 
ney  are  obvious  and  need  no  examples. 

2.  In  the  sale  of  a  bill  of  exchange  of  1563  rials  plate,  the  rial 
plate  was  estimated  at  9f  cents  ;  how  much  per  cent,  was  the  rial 
below  par  and  how  much  the  loss  ? 

Ans  4A  per  cent.     $6'94|  the  loss. 

3.  If  the  piastre  be  valued  in  exchange  at  81  cents,  what  is  the. 
percent,  above  par  on  a  bill  of  1672  piastres  5  rials  plate,  and  wha'. 
is  the  advance  on  the  bill  in  Federal  Money  ?  Ans. 

4.  OF  HAMBURGH. 

12  deniers=2  grotes  make     1  shilling  lubs,  or  stiver. 
lo  shilling  lubs=32  grotes      1  mark  banco-* 
3  <narks  1  rix  dollar. 

Dr,  12  grotes  or  pence  Flemish  make     1  shilling  Flemish. 
20  shillings  Fl.=7|^  marks  1  pound  Flemish. 

A  mark  is  |^  of  a  dollar,  or  33J-  cents  in  the  U.  Stales,  and  the 
Kix  dollar  is  equal  to  the  Spanish  dollar,  or  100  cents. 
The  mark  is  2|  shillings  Flemish. 

The  Bank  money  of  Hamburgh  is  superior  to  the  currency  at  a 
variable   rate  per  cent. 

1.  To  reduce  marks   banco  to  dollars,  divide  the  marks   by  3. 

3437 
Thus  3437  marks—— :^dolls.=g  1145  6C|cts. 

2.  To  reduce  pounds  Flemish  to  dollars,  multiply  the  pounds  by 
5,  and  divide  the  product  by  2  for  dollars.     Thus,  to  reduce  175 

175-5x5 
pounds  Fl.  and  10  shillings  to  dollars, —  dolls.=^438-75cts. 

3.  To  reduce  Hamburgh  money  to  sterling,  use  the  following 
proportion  ;  As,  the  value  of  a  pound  sterling  at  Hamburgh  is  to 
i  pound,  so  is  the  Hamburgh  sum  to  the  sterlmg  required. 

1.   When  the  pound  sterling  is  33  shillings  Flemish,  what  is  the 
value  of  £1567   10s.  Fl.  in  sterling  money  ? 
s.      £        £F\.      jSsterlin-. 
As  33  :  1  ::  1507-5  :  950 

*  Banco  is  money  placed  in  banks  of  deposit,  and  is  not  to  be  drawn  out,  buT. 
is  transferred  from  one  pcrroxi  to  another  for  the  payment  of  contracts. 


POLICIES  OF  INSURANCE.  297 

^.  Reduce  2560  marks  8  stii'ers  to  sterling,  at  the  rate  of  33i  shil- 
lingi  Fl.  per  pound  sterling.  Ans.  je20416-9  6d.  sterling. 

3.  When  the  pound  sterling  is  34  shillings  Flemish,  what  is  the 
per  cent,  below  par  ?  Ans.  4|  per  cent. 

4.  To  reduce  current  to  Bank  money,  use  the  following  propor- 
tion. As  100  marks  with  the  rate  added  is  to  100  bank  money,  so 
is  current  sum  to  the  bank  money  required. 

1.  Reduce  360  marks  current  to  bank  money,  when  rate  or  agio 
is  20  per  cent. 

As  100-4-20  :  100  ::  360  :  300  bank  money,  Ans. 

2.  When  the  rate  or  agio  is  18^  per  cent,  what  is  the  value  of 
3759  marks  8  stivers  current  in  bank  money  ?         Ans. 

3.  If  375  marks  current  are  estimated  al  320  marks  bank,  what 
is  the  rate  per  cent  ?  Ans. 

OF  CALCUTTA. 

12  pice  make  1  anna, 
16  annas  1  rupee. 

The  Bengal  rupee  is  estimated  at  50  cents  in  the  United  States  ; 
in  exchange  it  is  usually  3  or  4  cents  less. 

100  sicca  rupees  are  equal  to  116  current  rupees.     ' 

1.  Reduce  187  rupees  8  annas  to  federal  money  at  46i  cents 
per  rupee.  Ans.  §89  06i  cents. 

2.  Reduce  ^367^  to  rupees,  the  rupee  being  valued  at  48  cents. 

Ans.  763  rupees,  15  annas,  and  4  pice. 
Note.  From  the  exchange  value  of  the  money  of  different  coun- 
frie<^,  and  from  the  Table  of  Money  of  commercial  countries,  im- 
mediately before  the  "  Chronological  Problems,"  the  student  will 
be  able  to  derive  particular  rules  for  making  all  the  exchanges  ot* 
money,  which  may  be  necessary  in  business. 


POLICIES  OF  INSURANCE. 

INSURANCE  is  an  assurance  or  security  b^'  a  contract,  to  in- 
demnify, for  a  specified  sum,  the  insured  for  such  losses  as  the 
property  may  be  exposed  to,  for  a  certain  time. 

The  insurer  or  underwriter ^  is  the  party  that  is  bound  to  indem- 
nify for  the  loss  sustained. 

The  premiuni  is  the  compensation  paid  by  the  insured  for  the 
insurance. 

The  policy  is  the  document  by  which  the  contract  of  insurance  i^ 
made. 

Goods  are  said  to  be  covered,  when  their  value  and  the  premium 
and  other  charges  are  insured. 

If  the  loss  do  not  exceed  Jtu^r  per  cent,  the  underwriter  is  free, 
and  the  loss  is  borne  by  the  insured.  Particular  average,  is  the 
proportioning  of  such  lo.sscs  as  ari^<^  from  ordinary  acridents  atses, 

O  o 


29^  rOLlClES  OF  iiNSURAKCE. 

«tnong'  the  proprietors  of  the  property  which  suffers  the  injurj 
General  average^  is  the  proporUon  to  be  paid  by  all  the  owners  oi 
ship  and  cargo,  for  losses  necessary  to  preserve  the  rest,  such  as 
cutting  away  masts,  &c.  throwing  part  of  the  cargo  overboard,  and 
the  like.  As  this  is  done  for  the  common  good,  it  is  to  be  borne 
by  the  owners  of  the  ship  and  cargo,  in  proportion  to  the  value  ot 
the  property  possessed  by  them  severally. 

In  computing  general  average  for  masts,  &c.  to  replace  those 
cut  away,  one  third  is  usually  deducted  from  the  expense,  be- 
cause the  new  articles  may  be  supposed  better  than  the  old 

Unless  the  property  is  covered  the  insured  is  not  indemnified,  in 
case  of  total  loss,  but  in  the  proportion  contained  in  the  policy^ 
and,  in  case  of  a  partial  loss,  the  insdred  is  to  be  indemnified  only 
in  the  same  proportion. 

ISote.     General  average  is  computed  by  the  Rule  for  Single  Fel 
lowship.     See  examples  19  abd  20  under  that  rule. 

CASE  I. 

Wheti  the  premium^  at  a  certain  rate  per  cent,  for  insuring  a  suvi^  is 
requiredy  the  operation  is  the  same  as  in  interest^  or  couimission. 

Examples. 
1.  Wliat  is  the  premium  upon  5371.  15s.  9d.  at  6i  per  cent.  ? 

637  15     9 


^^ 

3226  14 
=  268  17 

6 

101 

34195  12 
'20 

4^ 

19)12 
12 

1|48 
4 

1|94  Ans.  £34  i 9s.  lid.  heaijj. 

2.  What  is  the  premium  upon  ^375,  at  7i  per  cent.  ? 

"  Ans.  $28*12o. 

CASE  II. 

To  find  the  sum  fur  which  a  policy  should  be  taken  out  to  cover  a 

given  sum. 
Rule.     Take  the  premium  from  1001.  or  jslOO,  and  say,  As  the 
remainder  is  to  100,  so  is  the  sum  adventured  to  the  policy.*     Or, 

*  It  is  plain,  that  the  policy  should  be  equal  to  the  insurance  and  the  sum  iu- 
sured.  Hence  at  8  per  cent,  ii  policy  of  £100  would  secure  only  £92.  In  order  to 
Recover  £92,  therefore, the  policy  must  be  taken  out  for  £100.     Henee  the  rule  is 


POLICIES  OF  INSURANCE.  299 

In  decimals,  take  the  premium  from  100,  annex  two  cyphers  to 
the  sum  to  be  corered,  and  divide  hy  the  remainder  for  the  policy. 

1.  It  is  required  to  coTcr  7591.  pren^iurq  8  percept.  :  For  what 
sum  must  the  policy  be  taken  ^     ' 

100 
8 

92  :   100  ::   759 
100 

£ 

02)75900(825  Ans. 
736 

230 
184 
-^ 75900 

460  Or, =£825,  Ans.  gis  before. 

460  92 

2.  A  merchant  sent  a  vessel  and  cargo  to  sea,  valued  at  JSTfjO : 
What  sum  must  the  policy  be  taken  out  for,  to  cover  this  property, 
premium  19i  per  cent.  ?  Ans.  ^7155  28c. 

CASE  III. 

When  a  policy  i$  taken  out  for  a  certain  sum  in  order  to  cover  a 
given  sum. 

To  find  the  premium,  say,  as  the  policy  is  to  the  covered  sum  ; 
so  is  1001.  (orglOO)  to  a  fourth  number,  which,  being  taken  from 
iOO,  will  leave  the  premium.     Qr, 

In  decimals,  diyide  the  sum  covered,  with  two  cyphers  annexed, 
by  the  policy  ;  subtract  the  quotient  from  100,  the  remainder  is  the 
premium. 

Example^. 
I.  If  a  policy  be  taken  out  for  12501.  to  cover  5001.  What  is  the 


premium  per  cent.  ? 


o>)vioUs.  The  difference  between  100  and  the  rate  per  cent,  will  be  the  first 
term,  100  the  second,  and  the  sum  to  be  insured  the  third  term  of  a  proportion, 
and  the  rule  is  merely  a  particular  application  of  the  Rule  of  Three.  In  the  first 
example,  the  proportion  would  stand  thus,  100 — 8  ;  100  ::  759  :  the  policy= 

-—- — ^-—-£825.    Now  the  premiun[i  on  £825,  is,  ——-—£66,   and   66-f- 

lOO — 8  100 

750=:£825,  tlie  policy.  The  rule  for  decimals  is  evidently  a  contraction  of  this 
rule. 

In  Case  III.  the  last  three  ternis  ip  the  preceding  proportion  are  given  to  find 
the  rate.  Those  three  terms  evidently  give  the  diflfererice  between  100  and  the 
rate,  and  the  rule  is  obvious. 

In  Case  IV.  the  first  two  terms  and  the  last  term  of  the  preceding  proportion 
are  given,  to  find  the  third  term  or  sum  covered,  and  the  reason  of  the  operation 
J5  \.l\in  from  tlio  con«idenilion  of  that  proportion. 


:?00  POLICIES  OF  INSURANCE 

1230  :   500  ::    100 
100 

1250)60000(40  and  £100— 40=:i:60,  Ads. 
50000 

Or, =40,  &c.  as  before. 

1250 
2.  It'a  policy  be  taken  out  for  g781-25,  to  cover  ^625  :  Requir- 
ed the  premium  per  cent.  ? 

*|>    c.         k  S       §    c. 

As  781-25  :  623  ::   100  :  87-30.     And,  100— 87-5=12-5,  or  121 
62500  [per  cent,  premiunn,  Ans. 

Or, =87-5,  &c.  as  before. 

781-25 

CASE  IV. 

When  the  policy  for  covering  any  sum  and  the  premium  per  cent,  are 
given,  tojind  the  sum  to  be  covered, 

RULK. 

Deduct  the  premium  per  cent,  from   100,  and  say,  As  IQO  is  to 
the  remainder,  so  is  the  policy  to  the  sum  required  to  be  covered. 
Or,     In  decimals^  Multiply  the  policy  by  the  remainder  found  a^ 
before,  and  point  off  two  right  hand  places  in  the  product  for  the  an- 
swer. 

Examples. 
1.  If  a  policy  be  taken  out  for  12501.  at  60  per  cent.  :    What  is 
the  adventure  or  sum  to  be  covered  ? 
100 
60 


100  :  40  ::    1250  Or,  1250x100—60=50000,  and,  < 

40  pointing  off  two  places,  500  00  ^ 

£  Ans.  as  before. 

100)50000(500  Ans. 
2.  If  a  policy  be  taken  out  for  $781  25c.  at  V2.\  pe«-cent.  requir 

ed  the  sum  covered  ?  

781-25X100—12^ 

As  100  :  100—121  :;  781-25  : ^=$625,  Ans. 

100 

Or,  781  •25x100—12  5=62500  ;  and  625-00,  Ans.  as  before. 

CASE  V. 

When  a  given  sum.  is  adventured  several  voyages  round  from  one  place  : 
to  another^  either  at  the  same,  or  di^/ferent  risks,  from  place  to  place,' 
and  it  is  required  to  take  out  a  policy  for  such  a  sum  as  will  cover  \ 
the  adventure  all  round,  supposing  the  risk  out  and  home  to  he  equal  ' 
and  tantamount  to  thz  several  given  risks. 

Rule. 
1.  Raise  1001.  or  ^100  to  that  power  denoted  by  the  number  o{ 

risks,  and  multiply  the  said  power  by  the  sum  adventured,  (or  to 

be  covered)  fur  a  dividend. 


POLICIES  OF  INSURANCE.  301 

2.  Subtract  the  several  premiums,  each,  from   1001.  and  multi- 
ply the  several  remainders  continually  together  for   a  divisor,  and 
^     the  quotient,  arising  from  this  division,  will  give  the  policy  to  cov- 
I   er  the  adventure  the  voyage  round.* 

Example. 

A  merchant  adventured  glSOO  from  Boston  to  Philadelphia,  at  3 
per  cent,  from  thence  to  Guadaloupe,  at  4,  from  thence  to  Nantz,  at 
o,  and  from  thence  home  at  6  per  cent. ;     For  what  sum  niust  he 
take  out  a  policy  to  cover  his  adventure  the  vogage  round,  suppos 
ing  the  risk  to  be  equal  out  and  home,  and  tantamount  to  the  sev 
eral  given  risks  ? 

100     X     100    X     100  X     100x1500 

-  ■ :.:== ..-    =g  1803-835,  Ans. 

100— 3X100— 4X100— 5X100--6         "^ 

CASE  VI. 

When  a  given  sum  is  adventured  several  voyages  round,  as  in  the  last 
case^  either  at  the  same^  or  different  risks,  from  port  to  port,  and 
the  premium  for  the.  voyage  round  is  required,  tantamount  to  the 
several  given  rates  per  cent. 

*  It  is  evident  that  the  policy  to  be  takeij  out  for  tlie  first  voyage  becomes  the 
sum  for  which  a  policy  is  to  be  taken  out  for  the  second  voyage,  and  so  on. 
Hence  the  examples  of  this  case  are  to  be  solved  by  the  rule  for  Case  II.  making 
the  sum  in  the  policy  for  the  fir^t  voyage,  the  sum  for  which  a  policy  is  to  be  tak- 
en out  for  the  second  voyage.  Therefore  the  operation  on  tlie  given  example 
would  be  as  follow*. 

100 — 3  :  100  ::  1500  :  policy  for  1st  voyage=: -—.     Now  as  — ^-- 

100 — 3  100 — 3 

is  the  sum  to  be  insured  on  the  second  voyage,  we  have, 

.nr.     .      inn        100x1500      ^    ,      ^.          100x100x1500 
100--4  :  100  ::  — —  -  ;  2nd  policy^. —  . 

^  ^"^'^  1 00— yj  X  J  00—4 

»    1  .nn     r      inn        100X100X1500      ^^      ^.  100X100X100X1500 

And  100 — 5  :   100  ::  zrriizr ■    ■  ■  ■  -  :  3d  policyrir:—  — — . 

1 00— 3  X 100—4  100—3  X 100—4  X 100— S 

.    ,    ^^     .      ^^         100X100X100X1500       ^^, 

And  100—6  :  100 ::  ■. :  4th  policy= 

100—3  X 100—4  X  100—3 
100X100X100X100X1500 1004X1500 

100—3  X 100— 4  X  Too— 5  xToO— 6        100^3  X  100—4  X  100— ^X  100— 6 
which  is  the  Rule.     The  same  may  be  shown  by  the  Double  Rule  of  Three,  thas, 
100—3  :  100  ::  1500  :  V,         ,.  100*  X1500 

100—4  :  100  ::  :    I  ^^^"^  polxcy--=  ._ ,: _     ^ 

100—5  •  100  ••  •    (  100—3  X  100—4  X  100—5  X  100— 1> 

100—6  :  100  ::  :  j  ==$1803  83c.  5m. 

It  is  plain  that  hov/ever  numerous  the  voyages,  the  power  of  100  must  be  equal 

to  their  number,  and  that  the  divisor  must  always  be  the  continued  product  of 

the  differences  between  100  and  the  several  rates  of  insurance.     If  tlie  rate  of  in- 

,     ,,_        .,  1004X1500 

snraace  had  been  the  same  on  each  of  the  voyo^'cs,  then  the  p-,\'-.-  - .— _  — 

li  the  rate  had  been  6  per  cc-nt. 


30S  POLICIES  OF  INSURANCE. 

Rule.* 

1.  Find  the  sum  for  which  the  policy  must  be  taken,  by  the  last 
case. 

2.  Multiply  the  sum  adventured  by  J 00,  and  divide  that  product 
by  the  policy. 

3.  Take  the  quotient  from  100,  and  the  remainder  will  be  the 
premium  per  cent,  on  the  policy,  tantamount  to  the  several  pre- 
miums given  ID  the  question. 

Example. 

A  merchant  adventured  ^1500  from  Boston  to  Philadelphia,  at 
3  per  cent.:  from  thence  to  Guadaloupe,  at  4;  thence  to 
Nantz,  at  5;  and  thence  home,  at  6  percent.:  What  will  b^ 
the  premium,  tantamount  to  those  given  in  the  question,  on  a  poli- 
cy for  covering  the  first  adventure,  the  whole  voyage,  supposing 
the  risks  out  and  home  equal  ? 

In  Case  V.  we  found  the  policy,  which  would  cover  the  adventure 

1500x100 
the  vo-yage  round,  to  be  <J  1803  835.     Then   100 —  TwF835^ 

16'844=;the  premium  per  cent,  on  the  policy  the  voyage  roundj, 
aad  tantamount  to  the  several  given  premiuajs. 

CASE  VII. 

If  a  policy  be  taken  out  for  a  given  surriy  to  cover  a  certain  advefiiurc^ 
from  one  port  to  another^  on  to  several  ports,  at  equal  premiums 
from  one  place  to  the  other ^  to  find  what  that  equal  premium  is. 

RuLE.t 

.1.  Involve  100  to  that  power  denoted  by  the  number  of  risks^ 
and  multiply  this  power  by  the  sum  adventured,  (or  covered.) 

*  When  the  policy  is  found  by  Case  V,  the  operation  becomes  the  same  as 
tliat  directed  by  the  Rul«,  Ct^s*  III.  which  has  been  proved.     The  operations 
pray  be   shortened  in   mnny  cases,  by  keeping    the  terms  separate  in  Ihe 
6r4  part  of  the  process.     Thus— by  C^e  V.  thp  policy  in  tUis  examples- 
loo*  x  1500 

Then,  by  Cage  II. 


100— 3  X  100— 4  X  100—5  X  100- 
1004  X1500 


•  1500  ••  100  •    ^  ^^^  diminished  l^y 


iOO— .iX  100— 4  X  100—5  X  100—6    _    ^ (      the  premuimr 

1500  X 100  X  ioO— 3X  100—4  X  100— .Tx  100—6 
l(Kr43<ri7>00 


^O0^^X10O-4X1O()-r.x  ■<X^^^^33.t,6^  ,„a  100-03-156=$16-S44. 

!•  By  llip.  Iii?t  remark  in  the  demonstration  of  the  Rule  CaseA^  -when  the  in- 
auranoe  is  the  feame  on  each  of  several  voyages,  the  poJkj/  is  equal  to  the  pro- 
duet  of  the  sum  to  be  insured  and  100  raised  to  a  power  whose  index  is  the 
number  of  voyages,  divided  by  the  difference  between  100  and  the  rate  of  in- 
iuvance  r.iised  to  the  same  power.  Hence  this  product  divided  by  the  poli'  • 
aplfA  5;ive  a  quolieut  ciun.1  to  the  c'Ceron'^o  between  100  end  the  rc.te  rf  ins'^> 


POLICIES  OF  INSURANCE.  Si^j 

i.  Divide  the  last  product  by  the  policy. 

3.  Extract  that  root  of  the  quotient  denoted  by  the  number  of 
risks. 

4.  Take  this  root  from  100,  and  the  remainder  will  be  the  equal 
premium  from  one  port  to  the  other. 

Example. 
A  merchant  adventured  <J1500  from  Boston  to  Philadelphia, 
thence  to  Guadaloupe,  thence  to  Nantz,  and  thence  home  ;  to  co- 
ver which  all  round  he  took  out  a  policy  for  ^1803  635  ;  and  the 
premium  was  equal  from  one  place  to  the  other  :  what  was  th« 
premium  per  cent.  ? 

,r^^       ^lOOxlOOXlOOx  lOOx  1500     ^  ^^^  ,    . 

100—*/ C: C _U — =4-507  per  cent.  Answer. 

1803-835 

CASk  VIII. 

When  an  adventure  is  insured  out  and  home  at  one  risk,  at  n  ^ivcn 
rate  per  cent,  and  the  voyage  terminalcs  short  of  what  Tvas  at  first 
intended  :   To  find  what  the  underzvritcr  must  receive  per  cent. 

Rule. 

1.  If  just  half  the  voyage  is  performed,  it  must  be  consideretf 
as  two  equal  risks :  If  one  third,  then,  as  three  equal  rij^ks ;  if  buf 
one  fourth,  then,  as  four  risks,  and  so  on  ;  and  by  Case  2d  mu*t 
be  found  the  amount  which  will  cover  the  adventure  the  voyage* 
round. 

2.  Involve  100  to  that  power  denoted  by  the  number  of  risks, 
and  multiply  this  power  by  the  sum  adventured. 

3.  Divide  this  product  by  the  aforesaid  amount. 

4.  Extract  that  root  of  the  quotient  denoted  by  the  number  ot 
risks. 

5.  Take  this  root  from  100,  and  the  remainder  will  be  tlie  sum 
per  cent,  which  the  underwriter  must  receive. 

Example. 

A  merchant  covers  §200  at  6  per  cent,  from  Newburyport  to 
the  West  Indies  and  home  again  ;  but  the  voyage  terminating  in 
the  West  Indies,  what  must  the  insurer  receive  jver  cent.  ? 
100 

6 

94  :  100  :;  200  :  212-765957=amount  to  cover  §200  voyage  round, 

2000000  ^ 

100X100X200=2000000  and  ^jpj^— =9400,  and  100~v'9400 

—30465  to  be  paid  the  insurer  per  cent,  upon  the  above  amount, 

ance  raised  to  a  power  whose  index  is  the  number  of  years.  U  that  root  of  th;^ 
quotient,  indicated  by  the  number  of  year^,  be  extracted  you  will  have  the  dif- 
ference between  100  and  the  rate  per  cent,  and  this  differen/"?  takn  frona  10<:* 
sires  the  rate. 


304  Compound  interest. 

COMPOUND  INTEREST 

IS  that  which  arises  from  the  interest  being  added  to  thl 
ripal,  and  (continuing  in  the  hands  of  the  borrower)  becoming 
part  of  the  principal,  at  the  end  of  each  stated  time  of  payment. 

Method  I. 

Rule.* — Find  the  amount  of  the  given  principal,  for  the  time  of 
the  first  payment,  by  Simple  Interest :  next,  find  the  interest  of 
that  sum,  or  principal,  and  add  it  as  before,  and  thus  proceed  for 
any  number  of  years,  still  accounting  the  last  amount  as  the  prin- 
cipal for  the  next  payment.  The  given  principal  being  subtract- 
ed from  the  last  amount,  the  remainder  will  be  the  compound  in- 
terest. 

In  federal  money,  multiply  the  principal  by  the  rate  for  the  first 
time  of  payment,  setting  the  product  two  places  more  to  the  right 
than  the  multiplicand,  and  the  decimal  point  in  the  product  under 
that  in  the  multiplicand  ;  therv  find  the  amount,  and  proceed  as 
above. 

A''otc.  It  is  not  usually  necessary  to  carry  the  work  beyond  mills  ; 
therefore,  when  the  figure  next  beyond  mills,  at  the  right,  exceed:^ 
5,  increase  the  number  of  mills  1  ;  when  it  does  not  exceed  5,  it 
may  be  omitted.  The  result  will  be  exact  enough  for  commoa 
purposes. 

Examples. 
I.  What  will  £480  amount  to  in  5  years,  at  6  per  cent,  per  an- 
num ?  £ 

Principal  480  Principal  for  the  1st  year  480     0 

Rate  of  interest  G  Interest  of  ditto    28   16 


S8|80  Principal  for  the  2d  year  508   16 

20  6 


16J00  30|52   16 

£     s.  d.  20 

Prin.  for  the  2d  year  508   16  0 


Interest  for  ditto     30   10  6} 


Prin.  for  the  3d  year  539     6  6 


32|35   19  3 


6  6 1 72 

4 


20  2j88 
Carried  up. 

"^  It  may  be  observed  that  all  the  computations,  relaluij^  to  CompoimJ  hdcr 
est,  are  founded  upon  a  series  of  terms,  increasin^:  in  Geometrical  Progression, 
■wherein  the  number  of  years  assigns  the  index  of  the  last  and  highest  term  - 
Therefore,  as  one  pound  is  to  the  amount  of  one  pound,  for  any  given  time.  ;  ( 
i-  any  proposed  principal,  or  sum.  to  it^a  amount  for  the  same  time. 


COMPOUND  INTEREST.  305 

Brought  up.         1|97     Principal  for  the  3d  year £539     6  6^ 
12  Interest  for  ditto     32     7  2-J 


2(31     Principal  for  the  4th  year    571  13  8^ 

4 

1|24                                                      £  s.    (1. 

£     s.    d.  Prin.  for  the  4th  year  571  13  8^ 

Prin.  for  the  4th  year  671    13  8£-         Interest  for  ditto    34  6  0]- 


o                                                       — 
—  Prin.  for  the  5th  year  605  19  9 

34130  2 
20 

4}                                                     (> 

36)35   18  6 

6|02 

20 

12 

7]18 

o;28 

12 

4 

■ — 1 — . 

2'22 

1)14 

• 

£     s.     d. 

iTincipal  for  the  5th  year  605  19     9 

Interest  for  ditto     36     7     2 


Amount  for  5  years  642     6   11 
v'^nbtract  the  first  principal  480     0    0 

Compound  interest  for  5  years  162     6   11 
'n  federal  money,  thus  :   The  principal  beinsT  §1600 for  five  ycaps, 
Principal  for  the  1st  year  §1600- 
Rate  of  interest    6 


Interest  1st  year      96  00 

Amount  1st  and  prin.  2d  year  169d- 

6 

Interest  2d  year     101-76 

Amount  2d  year,  prin.  3d  1797*76 

6 


Interest  3d  year     107-8656 

Araouiit  3d,  principal  4th   1905-6256 

6 


Interest  4th  year     1 14-33753t 


Amount  4th,  principal  5th  ve^r  2019  963136 

6 
P  p  Carried  over. 


30G  COMPOUND  INTEREST. 

Brought  over.  Interest  5th  year     121'197788}6 


Amount  for  5  years  214M6092416 
Subtract  Ist  principal  1600- 

Compound  Interest  for  5  years=     541-16092416 
Or  thus : 
1st  principal  J 1600* 
6 


Interest     96-00 


2d  principal  1696- 
6 


Interest     101-76 


3d  principal  1797-76 
6 


Interest     107-866 


4th  principal   1905626 
6 


Interest     114-338 


6th  principal  2019-964 
6 


Interest     121-198 


Amount  214 M62 
Jst  principal   1600" 

Compound  Interest     541-162  nearly,  as  before. 

2.  What  is  the  compound  interest  of  ^740  for  6  years,  at  4  per 
cent,  per  annum?  Ans.  ^196  33c.  6m. 

3.  What  will  £400  amount  to  in  5  years,  at  £4  per  cent,  per 
annum?  Ans.  £48^  13s.  2id. 

4.  What  will  £150  amount  to  in  a  year,  at  2  *per    cent,  per^ 
month?  Ans.  £190  4s.  5d. 

5    What  is  the  compound  interest  of  ^500  at  2  per  cent,  a  month^ 
for  one  year?  Ans.  $134  12c.  Im. 

6.  What  is  the  amount  of  glOO  at  6  per  cent,  compound  inter- 
est for  3  years  ? 

7.  What  is  the  compound  interest  of  g  100  at  7  per  cent,  for  3 
years? 


COMPOUND  INTEREST.  307 

Methop  II. 

When  the  rate  is  at  5  per  cent,  per  annum. 

1.  Divide  the  principal  by  20,  and  this  quotient,  added  to  the 
principal,  will  be  the  amount  for  the  iirst  year,  and  the  principal 
Tor  the  second. 

2.  In  like  manner  find  the  amount  for  every  succeeding  year. 

}Vheji  the  rate  is  at  6  per  cent,  per  annum. 

1.  Divide  the  principal  by  20,  and  that  quotient  by  5:  these 
quotients,  added  to  the  principal,  ivill  be  the  amount  for  the  first 
vear,  and  the  principal  for  the  second.  • 

2,  In  like  manner  obtain  the  amount  for  every  succeeding  year. 


Examples 

1.  What  is  the  amount  of £480 
at  6  per  cent,  per  annum,  for  5 


years  ? 
20)480 
5)  24 


4   10 


20)508   16  amount  of  1st  year. 

5)  25     8     91 
5     1      9 


J0)539     6     6i  ditto  of  2d. 
5)  26   19     3f 


7   lOJ- 


20)571    13     8^  ditto  of  3d. 


5}  28   11      8-1 
5   14     4 


20)605  19     83  ditto  of  4th. 
5)  30     5   llf- 
6      1      2i 


2.  Of  the  same  sum  at  6  per 
cent,  per  ^nnum,  for  5  years. 

£ 

20)480 
24 


20)504        amount  of  1st  year 
25     4 


20)529     4  ditto  of  2d. 

26     9     2|- 


20)555  13     21 
27   15     7| 


20)683     8   JO     ditto  of  4th. 


29     3     51- 


£612   12     31  do.  of  6th.  Ans, 

JVote.  The  same  may  be  done 
in  federal  money,  but  the  first 
method  is  generally  more  easy. 


.0612     6   10|  do.of5tb.  Ans. 

COMPOUND  INTEREST  BY  DECIMALS. 


'\  Table  of  tlxo  Amount  of  J^l  or  ^],  at  -^  per  cent,  per  month,  as  practised  at 

the  Banks. 


Months. 

Dec.  parts- 

— r- 

Months. 

Jt:  or  $ 
Dec.  parts. 

Months. 

Ji  or  S 
Dec.  parts. 

1  I 

! 

1  005 
1  01 
1  015 
I  02 

5 

6 

7 
8 

1025 
103 
1035 
104 

9 
10 
11 

12 

1  045 
105 
1  055 
106 

308 


COMPOUND  INTEREST  Bt  DECIMALS. 


A  Table  of  the  Amount  of  £1  cr  $\,  from  1  Day  to  31  Days,  at  6  per  cent 

per  annum. 


... 
Days. 

Dec.  parts. 

Days. 

£  or  $ 
Dec.  parts. 

Days. 

£  or  $  • 
Dec.  parts. 

1 

1-00016 

12 

100197 

22 

1  00361 

2 

1-00032 

13  ' 

1-00213 

23 

1  00378 

3 

1-00049 

14 

1-0023 

24 

1-00394 

4 

1  00065 

15 

1  -00246 

25 

1-0041 

5 

1  -00082 

16 

1-00263 

26 

1  -00427 

6 

1-00098 

17 

1  00279 

27 

1- 00443  1 

7 

1-00113 

18 

1-00295 

28 

10046 

8 

1-00131 

19 

1  003 12 

29 

1-00476 

9 

1  00147 

20 

1-00328 

30 

1  00493 

10 

1-00164 

21 

1-00345 

31 

1  00509 

11 

1-00180 

These  tables  are  formed  by  adding  the  interest  of  £1  or  ^1,  to 
£  1  or  $1,  for  the  given  rate  and  tinne.  Thus,  hy  rule  for  Simple 
Interest,  the  interest  of  £1  or  $1  fur  1  day,  is,  -00016438-1-,  and 
the  amount  is  100016438-f . 

CASE  I.* 

When  the  principal^  the  rate  of  interest,  and  time,  are  given,  to  Jin d 
either  the  amount  or  interest. 

Rule. 

1.  Find  the  aniount  of  £1  or  gl  for  one  year  at  the  given  rate 
per  cent. 

2.  Involve  the  amount,  thus  found,  to  such  power,  as  is  denoted 
by  the  number  of  years  ;  or,  in  Table  i.  at  the  end  of  Annuities, 

*  The  reason  of  the  rule  may  be  seen  by  the  following  process.  If  the  rate 
be  -6  per  cent,  the  amount  of  £l  or  $1  for  1  year,  is,  by  the  rule  for  Simple  In- 
tere&t  by  Decimals,  r06.     This  is  the  principal  for  the  second  year,  and  its 

amount  is  by  the  same  rule,  1-06-f  1-06  X  '00=^4^06  X  l-0G=l-06  X  VOe^Toy" . 
That  is,  the  amount  of  £1  or  $\  for  tioo  years  is  equal  to  the  square,  of  the 
amount  of  £l  or  $i  ior  one  year.     This  is  tlie  principal  for  the  third  year,  and 

its  amount  is,  VOtJ  ""-f  Tos"  X  •06=l-i-'06  X  ^0(7"=  1-06  X  K)6  =1-06  ,  thai  is 
the  amount  for  three  years  is  the  cube  of  the  amount  for  1  year.  In  the  Kanie  Avay 
it  maybe  shown,  that  the  amount  ior  four  years  is  the  /<mr//A  power  of  the 
amouat  of  £l  or  .^1  for  1  year;  ior  Jive  years,  is  the  Jiflh  power,  and  so  on. 
The  same  would  be  true,  whatever  be  the  rate  per  cent.  Now,  whatever  be 
the  principal,  the  amount  must  be  so  much  greater  than  the  amount  of  £l  or 
<S1  for  the  same  time  and  rate.  Therefore,  the  amount  for  any  })rincipal  will  he 
found  by  multiplying  the  amount  of  £l  or  ^1,  at  the  given  rate  and  time,  by  the 
principal  and  is  the  rule.     Let  the  principal  be  5^00  or  £100,  the  rate  5  P'^r 

,       ,'   '  5  — . —  S 

cent,  and  the  time  5  years,  ^ben,  i-Oa  XlO0=Uie.  amount.  And  1-05  X 
100— 100=the  interest. 

If.the  rate  of  interest  be  dcterr^iined  to  any  other  time  than  a  year,  as  ^,  J,  Ic. 
the  rule  i,?  the  same. 

If  the  compound  interestyor  amount  of  any  sun>,  be  required  for  tlie  pailg  of 
ft  A'ca'r,  it  nxixy  be  deteraiincd  as  foliuAj's  : 


COMPOUND  INTEREST  BY  DECIMALS.  309 

under  the  rate,  and  against  the  given  number  of  years,  you  will 
lind  the  power.* 

3.  Multiply  this  power  by  the  principal,  or  given  sum,  and  the 
product  will  be  the  amount  required,  from  which  if  you  subtract 
the  principal,  the  remainder  will  be  the  interest. 

Examples. 

1.  What  is  the  compound  interest  of  £600  for  4  years,  at  6  per 
cent,  per  annum  ?  C  amount  of  £l  for  1  year,  at  6  per 

-,-.,-      ^  ^„      i  cent,  per  annum. 

Multiply  by  1*06      ^  ^ 


M236=2d  power. 
Multiply  by  1-1236 

l-26247696=4lh  power- 
Multiply  by  COO=principal. 

757-48617600~amount. 
Subtract  600 


157-486176=£157  9s.  8id.=interest  required. 
By  Table   I. 
Tabular  amnt. of  £1  for 4 years,  at  6  percent. per  ann.=l '2624760 

Multiply  by  the  principal=  600 

Amount=757-4861400 
2.  What  is  the  amount  of  gl500  for  12  years,  at  3i  per  cent, 
per  annum  ? 

gl  0.35— amount  of  J^l  for  1  year  at  3^-  per  cent,  per  annum. 

And,  1035»2xl506=^2266  60c.  nearly,  Ans. 
Another  method  of  working  compound  interest  for  years^  months, 
Mid  days,  which  is  much  more  concise  than  the  preceding  method. 

I.  When  the  time  is  an  aliquot  part  of  a  year. 
Rule  1 .     Find  the  amount  of  £1  for  1  year,  us  before,  and  that  root  of  it, 
vvhich  is  denoted  by  \he  aliquot  part,  will  be  the  amount  of  j£l  for  the  time 
sought. 

2.  Multiply  the  amount,  thus  found,  by  the  principal,  and  it  will  be  the  amount 
of  the  given  sum  required. 

11.  When  the  time  is  not  an  aliquot  part  of  a  year. 
RuT.E  1.     Reduce  the  time  into  days,  and  the  3t)5th  root  of  the  amount  of 
iJl  for  1  year  is  the  amoiuit  for  1  day. 

2.  Raise  this  amount  to  that  power,  whose  index  is  equal  to  the  number  of 
days,  and  it  will  be  the  aniovmt  of  £l  for  the  given  time. 

3.  Multiply  this  amount  by  the  principal,  and  it  will  be  the  amount  of  the 
given  sum  required. 

*  Tire  amounts  of  £l  or  $1  in  this  treble,  are  so  many  powers  of  the  amount 
of  £1  or  $1  for  1  year ;  whose  indices  are  denoted  by  the  number  of  years. 

Note.  When  the  given  time  consists  of  years  and  months,  or  years,  months, 
and  days  ;  first  seek  the  amount  of  £l  or  $1  in  the  table  of  years,  then  in  the 
table  of  months,  &c.  multiply  these  several  amounts  and  the  principal  continu- 
ally together,  and  the  last  product  will  be  the  amount  required. 

Thus,  if  the  amount  of  £480  in  5^  years,  at  6  per  cent,  per  annum,  were  re- 
quired ;  the  amount  of  £l  for  5  vears=£  1-33822,  dittc  for  6  months— £l "0:2900. 
Now,  l-33822Xl-029G6x4U0=:£66i-2341  Answr-r. 


^iO 


COMPOUND  INTEREST  BY  DECIMALS. 


Rule. 
To  the  logarithm  of  the  principal,  found  in  any  Table  of  loga- 
rithms, add  the  several  logarithms,  answering  to  the  number  of 
years,  months  and  days  found  in  the  following  tables,  and  their  sum 
tviH  be  the  logarithm  of  the  amount  for  tlie  given  lime,  which  be- 
ing found  in  any  table  of  logarithms,  the  natural  number  corres- 
ponding thereto  will  be  the  answer.* 


iOCARITHMICK  TABLES, 


SIX  I'ER  CEJVT.  PER  ANNUM,  FOR  YEARS,  MONTHS 
AND  DAYS. 


Years. 

Dec.  pis. 

Y. 

Dec.  pts. 

Y.' 

Dec.  pts. 

Y. 

Dec.  pts. 

Months 

Dec.  pts. 

1 

•025306 

n 

•278366 

21 

•531426 

:il 

-784586 

1 

•002166 

1   2 

•050612 

12 

•303672 

22 

•556732 

32 

•809792 

2 

•004321 

3 

•075918 

13 

•328978 

23 

•582038 

33 

-835098 

3 

-006466 

4 

'  -101224 

14 

■354284 

24 

•607344 

34 

•860404 

4 

•0086 

5 

•12653 

15 

•37969 

26 

•63265 

35 

•88571 

5 

•010724 

6 

1  -151836 

16 

•404896 

26 

•657956 

,36 

•911016 

6 

•012837 

7 

-177142 

17 

•430202 

27 

•683262 

37 

•936322 

7 

•01494 

8 

•202448 

18 

•455508 

28 

•708568 

■38 

•961628 

8 

•017033 

9 

-227754 

IG 

•480814 

29 

•733974 

39 

•986934 

9 

•019116 

10 

•25306 

20 

•50612 

30 

•75938 

40 

1^01224 

10 
11 

•021189 
•023252  ! 

Days. 

D. 

D. 

14 

D. 

20 

D. 

! 

1 

•000071 

•000571 

•000999 

•001426 

26 

•001852 

2 

•000143 

9 

•000642 

15 

•00107 

21 

•001497 

27 

•001923 

3 

•000215  |10 

•000713 

16 

•001142^22 

•001568 

28 

•001994 

4 

•000287 

11 

•000785 

17 

•001213 

23 

•001639 

29 

•002065 

5 

•000358 

12 

•000857 

18 

•001284 

24 

•00171 

30 

•002136 

1   6 

•000429 

13 

•000928 

19 

-CO  1355 

^5 

•001781 

31 

•00.2207 

1   7 

i 

•0005 

__. 

What  is  the  amount  of  1321.  10s. 
C^  years,  8  months,  and  15  da^s  ? 


at  6  per  cent,  per  annum, 


To  the  log 
f^oir.   for 


of  £132-5 
9   years 
Add  (  ditto  for  8  months 
(ditto    for   15  days 


=212221G 
=  •227764 
=  -017033 
=   -00107 

2-3t]8073 


per 


(cause  8  month?  are  pai?*,  deduct!  )  __.   .rjrynn.ioo 
cerjt.  upon  the  logarithm 'f  15  dajs  3 


Remains  23G80302,  the  nearer! 
to  which.,  to  the  table  of  logarilhais,  U  2  368101,  and  the  natural 
number  answerios:  thereto  is  233-4=  X233  8s.  Ans. 


i 


*  Aitt'ion^'li  tlirro  i*  a  small  erroi.fr  in  the  logarithm  for  days,  yet  they  are  ex- 
act enough  for  common  use.  And  if  after  the  first  month  you  deduct  ^  per  cent. 
•(.^^  each  mouth  past  (that  is,  ^  per  cent,  after  1  month,  H  per  cfent.  after  3 
/u;nithsT  fee.)  frojn  the  logarithm  of  tlie  number  of  days,  it  will  give  ^iie  true  an- 
fcW(.r. 

Wo'fe,  That,  after  1  mouth,  ^  per  cent,  on  the  logaritlim  of  1  day  is  -000000355, 
iyn  2  day?M,  is  -000090710  :  After  2  months,  1  per  cent,  on  the  ]ogaritl»m  of  1  dav, 
}.-< '-00000071, on  Sday^^,  -00000143  :"  After  10 iuonths,  5  per  cent.  '>u  the  lo^rarithin 
ri;r  1  day,  is  -(XKiOOSiy,  on  6  day?,  is,  -0000^^145,  &-•. 


COMPOUND  INTEREST  BV  DECIMALS. 

CASE  II. 
Whtn  the  amonnt,  rate  and  time^  are  givcriy  to  find  theprincipaL 

Rui-E. 

Divide  the  amount  by  the  amount  of£l  or  gl  for  the  given  tim-e, 
and  the  quotient  will  be  the  principal.* 

Or,  If  you  multiply  the  present  value  of£l  or  §1  for  the  given 
number  of  years,  at  the  given  rate  per  cent,  by  the  amount,  the 
product  will  be  the  principal  or  present  worth,  j 

Examples. 

1.  Wtiat  is  the  present  worth  of  7571.  9s.  8-}d.  due  4  years  hence, 
dsscounting  at  the  rate  of  61.  per  cent,  per  annum  ? 

By  Table  I. 

Divide    by  the     tabular  ^  _,   onnAnaa^^r'^    ioni4r,nr-PCTin    4«« 

.    rH    r      4  ^  =j*2G2476y)7a/-486i400(i/6OO  Ar». 

amount  ot  II.  tor  4  years,  S 

By  Table  11. 

Mult,  by  the  present  worth  of  11.  }  "^""^^^"^ -920936 
for  4  years,  at  C  per  cent  per  ann.  ^  _'         _^ 

Ans.  599-999923582704+=J1:60G; 

2.  What  principal  must  be  put  to  interest  G  years,  at  51  per  ct, 
per  annum,  to  amount  to  $689-4214033809453125  ?        Ans.  J!500. 

CASE  III. 

When  the  principal^  rate  andameunt,  are  gizerij  to  find  the  time. 

Rule. 

Divide  the  amount  by  the  principal:  then  divide  thfs  quotient 
by  the  amount  of  £1  or  $1  for  I  year,  this  quotient  by  the  same, 
till  nothing  remain,  and  the  number  of  the  divisions  will  show  the 
time. I 

Or,  Divide  the  amount  by  the  principal,  and  the  quotient  will  be 
the  amount  of  £l  or  gl  for  the  given  time,  which  seek  under  the 
given  rate  in  Table  1,  and,  in  a  line  with  it,  you  will  see  the  time. 

*  By  Case  I.  the  amount  is  equal  to  the  prh>cipal  multiplied  by  that  power  o{ 
the  amount  of  jSl  or  $1  for  1  year  at  the  given  rate,  which  is  indicated  by  the 
number  of  years :  therefore,  if  the  amount  be  divided  by  this  power  of  the  amount; 
cf  j£l  or  i^l  for  1  year,  the  quotient  must  be  the  principal.  Thus,  in  the  exam- 
ple in  the  proof  of  Case  I.  l*  05    XlOO:=:the  amount;  therefore,  '——■ 

1-05  •' 
100,  the  principal. 

t  See  Table  II.  shewing  the  present  value  of  jGl,  discounting  at  the  rates  of  4^, 
4j^,  tSzic.  per  cent,  the  construction  of  which  is  thus  : 
Amount.  Pres.  worth.  Amount.  Pres.  worth. 

As  1-06  :  1  ::  1  :  -9433962,  and  so  on,  for  any  other  rate  pe^ 
■eut.  and  time. 

.     I  By  the  example  in  the  proof  of  Case  L  VOo^  X 100— the  amount ;  divide  Ihif 
by  the  principal,  100,  and  the  quotient  will  be  1^^ .     This  quotient  divided  Iv 
the  ratio,  and  this  quotient  by  the  ratio,  and  so  on,  v.^ill  be  exhamte.l  by  fix-p  d  ■ 
^i'?ion«,  which  shows  th^  number  of  years. 


312 


DISCOUNT  BY  COMPOUND  INTEREST. 


Example. 
In  what  time  will  g500  amount  to  $689  42c.  Im.-f-,  at  6i  per 
cent,  per  annum  ? 


500 
'1-056 
1-056 
1065 
1-056 
1-056 
1-056 

689-421-f 
lvJ79-~ 

m 

a 

1-307-- 

1  -239— 

> 
-5 

M74-I- 

<D 

1113— 

1053-f- 

^ 


1-- 


Ans.  6 


years. 


CASE  IV. 

When  the  principal  i  amount  and  time^  are  given,  to  find  the  rate  per  it. 

Rule. 

Divide  the  amount  by  the  principal,  and  the  quotient  will  be  the 
j>mount  of  11.  or  %\  for  the  given  time  ;  then,  extract  such  root  as 
the  time  denotes,  and  that  root  will  be  the  amount  of  11.  or  gl  for 
1  year,  from  which  subtract  unity,  and  the  remainder  will  be  the 
ratio.* 

Or,  Having  found  the  amount  of  11.  or  gl  for  the  time  as  above 
directed,  look  for  it  in  Table  1st,  even  with  the  given  time,  and  di- 
rectly over  the  amount  you  will  find  the  ratio. 

Example. 
At   what   rate    per   cent,    per  annum  will   ^600    amount   to 
^689-421 403-f-  in  6  years  ? 

*;89-4214034-  ^ 
TKF^ =  1-378843—;  and  V«  1378843^  =1-055.     Then 

i  055— l-=^055=ratio 
num,  Answer. 


Hence  the  rate  is  b\  per  cefnt.  per  an- 


DISCOUNT  BY  COMPOUND  L\TEREST.] 

The  sum,  or  debt  to  be  discounted,  the  time  and  rate,  given,  to  find  the 
present  worth. 

Rule.  Divide  the  debt  by  that  power  of  the  amount  of  11.  or  gl 
for  1  year,  denoted  by  the  time,  and  the  quotient  will  be  the  pres- 
ent worth,  which,  subtracted  from  the  debt,  will  leave  the  discount. 


*  Proceeding  as  in  the  preceding  demonstration,  and  extracting  that  root  of  the 
quotient,  which  is  shown  by  the  number  of  years,  we  have  the  amount  of  £1  or 
.^1  for  1  year.     From  this  subtract  1,  and  the  remainder  is  tho  ratio.     Thus  ih 

s    —  — r- 
the  preceding  example,  i/   1*05 =1-05,  and  1*05 — 1='05,  the  ratio. 

t  As  the  present  worth  is  such  a  principal,  as  at  the  given  rate  and  time,  would 
ainount  to  the  debt,  this  rule  must  be  the  same  as  that  of  Case  H.  of  Compound 
Interest,  the  principal  being  in  this  case  \he  present  worthy  and  the  amoiint  the 
sum  or<hhf.     Or,  By  C;ise  1.  of  Compound  Interest  by  Decimals,  tixe  amount  of 


ANNUITIES,  3,13 

Examples. 
1.  What  is  the  present  worth,  and  discount,  of  £600  due  3  years 
hence,  atJ£6  per  cent,  per  annum,  oompound  interest? 
DiFide  by  UO6|'=i.l9101)600'00000(503-7741=£503  15s.   5|d. 
present  worth,  and  £600— £503   15  5|— £96  4s.  6id.=di9COunt. 
600  600 

Or,  j-^^Y^=£  503  7741,  and  600—  ^.^^^^^  =  £96-2259. 

By  Table   II. 

In  this  Table,  corresponding  to  the  time  and  rate,  we  have 

•839619~pre?ent  worth  of  £l  for  the  time  and  rate. 
Multiply  by         600=debt,  or  principal. 

503-771400=present  worth  of  the  debt. 
?.  What  is  the  present  worth  of  £312  10s.  due  2  years  hence, 
at  4^  per  cent,  per  annum,  compound  interest? 

Ans.£236  3s.  3d.  2-97qrs. 
3.  What  ready  money  will  discharge  a  debt  of  g  1000  due  4  years 
hence,  at  ^5  per  cent,  per  annum,  compound  interest  ? 

Ans.  g822  70c.  2m, 


ANNUITIES. 

AN  Annuity  is  a  sum  of  money  payable  at  regular  periods,  for 
a  certain  time,  or  for  ever. 

Annuities  sometimes  depend  on  some  contingency,  as  the  life  or 
death  of  a  person,  and  the  annuity  is  then  said  to  be  contingent. 

Sometimes  annuities  are  not  to  commence  till  a  certain  number 
of  years  has  elapsed,  and  the  annuities  are  then  said  to  be  in  re- 
version. 

The  annuity  is  said  to  be  in  arrears,  when  the  debtor  keeps  it 
beyond  the  time  of  payment. 

The  present  worth  of  an  annuity  is  such  a  sum  as  being  now  laid 
out  at  interest,  would  exactly  pay  the  annuity  as  it  becomes  due, 
and  is  the  sum  which  must  be  given  for  the  annuity  if  it  be  paid  ^ 
at  its  commencement. 

11.  or  $1  for  any  number  of  years,  is  equal  to  that  power  of  the  amount  of  ll.  or 
$1  indicated  by  the  number  of  years.  Hence  if  the  amount  be  11.  or  $\  the  prin- 
cipal will  be  the  reciprocal  of  the  power  of  the  amount  of  11.  or  $1 
inaicated  by  thenu:ii\>er  of  yeaJs  ;  thus,  if  l=amount  at  6  per  cent,  far  4  years, 

then,  =:the  principal  which  will  produce  the  amount  at  the  rate  apid  time. 

10&4  r  r  r  ^ 

Therefore,  if  l=:the  sum  to  be  discounted  at  that  rate  and  time,  then  ■■-•;  ^  -'=  its 

l'0o4 

present  worth,  and  is  the  rule. 

Note,     The  present  worth  of  11.  or  $1  for  any  time  and  rate,  is  the  recipracal 

ef  the  aiaount  of  11.  cr  ^l  for  the  same  time. 


^14  AiNNUlTlES. 

The  amount  is  the  sum  of  the  annuities  for  the  time  it  has  been 
forborne,  with  the  interest  due  on  each. 

CASE  I. 

7^0  *Jincl  the  amount  of  an  annuity  at  Simple  Interest. 

Rule. 

Multiply  the  sum  of  the  natural  series  of  numbers,  1,  2,  3,  4, 
&c.  to  the  number  of  years  less  1,  by  the  interest  of  the  annuity 
for  one  year,  and  the  product  will  be  the  interest  which  is  due  ou 
the  annuity. 

Multiply  the  annuity  by  the  time,  and  the  sum  of  the  two  pro- 
ducts, will  be  the  amount.* 

Examples. 

1.  What  is  the  amount  of  an  annuity  of  £100  for  four  years^ 
computing  interest  at  6  per  cent.  ? 
1  -{-24-3=6,  sum  of  the  natural  series  to  the  number  of  years  less  1. 

61.  interest  of  annuity  for  1  year. 
6x6=361.  the  whole  interest. 
100x4=4001.  product  of  annuity  and  time. 

Ans,  4361.  amount. 

2.  If  a  pension  of  g20  be  continued  unpaid  for  six  years, what 
is  its  amount  at  6  and  7  per  cent.  ? 

Ans.  At  6  per  cent.  ^138.     At  7  per  cent.  $141. 

3.  If  an  annuity  of  ^20  to  be  paid  halfeach  half  year  is  forborne 
for  six  years  ;  what  is  its  amount  at  6  per  cent.  ? 

Ans.  $169  60c. 

4.  If  a  pension  of  £33  is  forborne  for  12  years,  at  7  per  cent, 
what  is  the  amount  ?  Ans. 

CASE  II. 

To  find  the  present  worth  of  an  annuity  at  Simple  filtered.  M 

Rule.  fl 

Let  the  present  worth  of  each  year  be  found  by  itself,  discount- 
ing from  the  time  it  is  due ;  then,  the  sum  of  all  these  will  be  the 
present  worth.! 

*  It  is  plain  that  upon  the  first  year's  annuity  there  will  be  due  so  many 
year's  interest,  as  the  given  number  of  years  less  one,  and  gradually  one  year 
less  upon  each  succeeding  year,  to  that  preceding  the  last,  which  has  but  one 
year's  interest,  and  the  last  bears  none.  •  There  i^,  therefore,  due  in  the  whole 
as  many  years'  interest  of  the  annuity  as  the  sum  of  the  series,  1,  2,  3,  &;c.  to 
the  number  of  years  diminished  one.  It  is  evident  then,  that  the  wliole  inter- 
est due  must  equal  this  sum  of  the  natural  series  multiplied  by  the  interest  for 
one  year  ;  and  that  the  amount  will  be  all  the  annuities  or  tlie  product  of  thq 
annuity  and  time  added  to  the  whole  interest.     This  is  the  rule. 

t  This  rule  depends  on  the  principles  of  discount.  The  annuity  may  be  con- 
sidered for  each  year,  as  a  debt,  due  1,  %  3,&;c.  years  hence,  wf  which  tlie  pres- 


ANNUITIES  OR  PENSIONS  IN  ARREARS,  &p,      315 

Examples. 

1.  Find  the  present  worth  of  an  annuity  of  glOO  continued  five 
years  at  six  per  cent. 

i 

As  106  :  100  ::  100  :  943396,  the  present  worth  for  I  year. 
112  :  100  ::  100  ;  89-2857,  2  years. 

1 18  :  100  ::  100  :  84-7457,  3  years. 

124  :  100  ::  100  :  806451,  4  years. 

130  :  100  ::'lOO  :  76-9230,  5  years. 

^425-9301,  present  wortli  required. 

2.  Find  the  present  tvorth  of  an  annuity  of  751.  continued  for  4 
years  at  7  per  cent.  ?  Ans. 

3.  What  is  the  present  worth  of  a  pension  of  ^20  to  be  continu- 
ed for  6  years,  at  6  per  cent.  ?  Ans. 

ANNUITIES  OR  PENSIONS.  IN  ARREARS,  AT  COMPOUND 
INTEREST. 

CASE  I. 

^Fu€U  the  annuity i  or  pension^  the  time  it  continues^  and  the  rate  per 
cent,  are  given,  to  find  the  amount. 

Rule  I.* 

1.  Make  1  the  first  term  of  a  Geometrical  Progression,  and  the 
amount  of  j£l  or  §1  for  1  year  at  the  given  rate  per  cent,  the 
ratio. 

2.  Carry  the  series  to  so  many  terms  as  the  number  of  years, 
and  find  its  sum. 


mt  worth  is  to  be  found.     Hence  the  sum  of  the  present  worth  for  the  several 

[rears,  must  be  the  present  worth  for  the  whole. 

This  rule  is  very  absurd  in  practice.  It  is  obvious  on  inspecting  the  operation 
of  'Ex.  1 .  that  the  difference  between  the  present  worth  of  the  several  years  is 
I'.ontinually  diminishing.  Whence,  after  a  certain  number  of  years,  the  present 
worth  of  an  annuity  of  $100  would  produce  more  than  $100  interest  in  one 
year,  which  is  greater  than  the  annuity  to  be  purchased. 

*  I.  From  the  nature  of  an  amiuity,  as  explained  in  the  proof  of  the  rule, 
Case  I.  of  Annuities  at  Simple  Interest,  there  is  due  one  year's  interest  less  than 
the  number  of  years  the  annuity  has  been  continued.  Now,  by  Case  I.  of  Com- 
pound Interest,  the  amount  of  JCI  or  $1  at  the  given  rate,  is  equal  to  that  pow- 
er of  the  amount  for  one  year,  which  is  indicated  by  the  number  of  years.  This 
amount  is  obtained  for  one  less  than  the  number  of  years,  by  forming  the  geo- 
metrical series  as  directed  in  the  Rule,  or  beginning  with  unity.  Thus  in" Ex.  1, 
the  series  is,  1,  1-06,  1*062,  1-063,  and  the  last  term  is  the  amount  of  J2l  or  SI 
for  one  less  than  four^  the  number  of  years.  The  sum  of  this  series  is  the 
amount  at  Compound  Interest,  of  an  annuity  of  £,\  or  $1  for  four  years.  The 
amount  of  any  other  annuity  for  the  same  time  and  rate,  will  be  as  much  great- 
er or  less,  as  the  annuity  is  greater  or  less  than  £1  or  $1,  that  is,  the  amount  of 
the  annuity  of  £l  or  $1  must  be  multiplied  by  the  annuity  to  obtain  its  amount. 
(Hence,  the  rule  is  manifestly  correct.     In  Ex,  ],  the  above  series  amounts,  by 

Prob.  HI.  of  Geometrical  Profession,  to  — —H-.  and  this  multiplied  by  the 


316  ANNUITIES  OR  PENSIONS  IN  ARREARS, 

3.  Multiply  the  sum  thus  found  by  the  given  annuity,  and  the 
product  will  be  the  amount  sought. 

Rule   II. 

Or,  multiply  the  amount  of  £l  or  ^1  for  1  year  into  itself  so 
many  times  as  there  are  years  less  by  1  ;  then  multiply  this  pro- 
duct by  the  annuity  ;  and  subtract  the  annuity  therefrom.  Lastly, 
divide  the  remainder  by  the  ratio  less  1,  and  the  quatient  will  be 
the  amount. 

Examples. 

1.  What  will  an  annuity  of  601.  per  annum,  payable  yearly, 
amount  to  in  4  years,  at  61.  per  cent.  ? 

First  Method. 

l4-l'06-|-F66|'4-T'06p=4-374616=sum. 

Multiply  by  60=annuily« 

262-476960 
20 

9-53920 
12 


6-4704 
4 


1-8816     Ans.£262  9s.  6id. 


Or,  l-fl-06-{-l-06r4-l-06'''x60==je262  9s.  6^-d, 

Second  Method. 
l-06Xl*06x  l-06xl-06  =  l-26247 

Multiply  by  60  annuity. 

75-74820 
Subtract  60 
Carried  up. 

i-OCi — I 

annuity,  GO,  gives  the  amount  required,^: X60=2b2-4769C. 

•06 

]-064 — 1 

II.  The  second  rule  is  derived  from  the  expression, x60  ;  for  it  i 

•06 
,       1-064x60—1x60      ,       ,  ,.     .         , 

also,  — =the  above  amount,  and  is  tlie  rule. 

*0o 

Because  the  amounts  of  annuities,  at  the  same  rate  and  for  tlie  same  time,  ar 
as  the  annuities,  if  the  amount  be  divided  by  the  amount  of  j£l  or  $1  for  the 
same  time  and  rate,  the  quotient  will  be  the  annuity.     This  is  the  2d  Rule 
under  Case  II.    And  the  2d  Rule  of  Case  III.  is  readily  inferred  from  the  same 
principle. 


AT  COMPOUND  INTEREST.  317 

Brought  up. 
Divide  by  1'06— l=-06)15'7482(262'47=262l.  9s.  4|d.  Ans, 
12 

37 
36 

14 
12 

28 
24 

42 

42 


1-06x1 -06x1  06X1 -06X60— 60 
Or, jTSel^ =£262.47. 

Or,  bv  Table  III.* 

Multiply  tbe  tabular  number  under  the  rate,  and  opposite  to  tht 
lime,  by  the  annuity,  and  the  product  will  be  the  amount. 

2.  What  will  an  annuity  of  601.  per  annum  amount  to  in  20  years, 
allowing  61.  percent,  compound  interest? 

Under  61.  per  cent,  and  opposite  20,  in  table  3d,  you  will  find, 
Tabular  number=36-78559 

Multiply  by  60=annuity. 

2207-13540=22071.  2s.   Bid.  Ans. 

3.  What  will  a  pension  of  §75  per  annum,  payable  yearly, 
amount  to  in  9  years  at  5  per  cent,  compound  interest  ? 

Ans.  $826  99  233^m. 

4.  If  a  salary  of  1001.  per  annum,  to  be  paid  yearly,  be  forborne 
5  years,  at  61.  per  cent.  What  is  the  amount  ?     Ans.  5631.  14s.  2d, 

6.  What  will  wages  of  $25  per  month,  amount  to  in  a  year,  at|' 
per  cent,  per  month  ?  Ans.  $308  38c.  9m. 

CASE  n. 

When  the  amount y  rate  per  cent,  and  time  are  given,  to  find  the  annuity ^ 
pension^  ^c. 

Rule  I. 
Multiply  the  whole  amount  by  the  amount  of  U.  or  $1  for  a  year, 
from  which  subtract  the  whole  amount,  divide  the  remainder  by 
that  power  of  the  amount  of  11.  or  $1  for  a  year,  signified  by  the 
number  of  years,  made  less  by  unity,  and  the  quotient  will  be  the 
answer. 

*  Table  3  is  calculated  thus :  Take  the  first  year's  amount,  which  is  11.  mul* 
tiply  it  by  l-06+l=2-06=:second  year's  amount,  which  also  multiply  by  1-06-f 
l:=:3-1836=third  year's  amount,  &:c.  and  iu  this  manner  proceed  in  calculating 
tables  at  any  other  rates. 


5iS  ANNUITIES  OR  PENSIONS  IN  AHREAIIS, 

Rule  II. 
Or,  imd  the  amount  of  an  annuity  of  11.  or  $1  for  the  given  time 
and  rate  (by  Case  1 ;)  divide  the  given  sum  by  this  amount  j  and 
the  quotient  will  be  the  annuity  required. 

Examples. 
1.     What   annuity,    being    forborne  4   years,    will    amount   to 
-€262*47696,  at  61.  per  cent,  compound  interest? 
262.47696=amount. 
Multiply  by  106=amount  of  H.  for  1  year. 

157486176  106 

262476960  106 


278-2255776  636 

Subtract  262-47696  1060 


=26247696)15-7486176fje60  Ans.  1-1236 

15-7486176'  106 


0  -67416 

112360 

1,191016 

262'47696xl -06— 262-47696  1-06 

Or, =60. 

l-06xl-06xl*06xl06— 1  7146096 

11910160 


1-26247696 
Subtract  1- 


Divisor=-26247696 
Or,  thus. 
Aioount  of  all  annuity  of  11.  for  4  years  at  6  per  cent,  per  annum 

26247696 
n;:4.374616  (by  Case  1  ;)   and   4.37461^  ^''^^Q  ^"^• 

Or,  by  Table  HI.  the  amount  of  U.  is  found  to  be  4374616  ;  and 
the  answer  is  found,  as  before. 

2.  What   annuity,   being    forborne    20  years,  will    amount   to 
^2207-1354,  at  6  per  cent,  compound  interest  ? 

Amount  of  an  annuity  of  ^1  for  20  years  at  6  per  cent,  per  an- 
T}um=^36'78559.     And, 

36-78559)2207-13540(g60,  Ans. 
2207  1354 


0 

CASE  HI. 

iVken  th€  annuity f  amount  and  ratio  arc  givcTiy  iofind  the  time. 

HULE    I. 

Multiply  the  amount  by  the  ratio,  to  this  product  add  the  annui- 
jY,  and  from  the  sum  subtract  the  amount ;  this  remainder  bein^ 


AT  COMPOUND  INTEREST.  319 

uivitled  by  the  annuity,  the  quotient  will  be  that  power  of  the  ra- 
tio signitied  by  the  time,  which  being  divided  by  the  amount  of  11. 
for  1  year,  and  this  quotient  by  the  same,  till  nothing  remain,  the 
number  of  those  divisions  will  be  equal  to  the  time.  Or,  look  for 
this  number  under  the  given  rate  in  table  1,  and  in  a  line  with  it, 
you  will  see  the  time.     Or, 

Rule  H. 

Divide  the  amount  by  the  annuity  ;  from  the  quotient  subtract  1  , 
from  the  remainder  subtract  the  ratio  ;  from  successive  remainders 
subtract  the  square,  cube,  &c.  of  the  ratio,  till  nothing  remain  ;  and 
the  whole  number  of  the  subtractions  will  be  the  answer.  Or,  find 
the  quotient  in  Table  III.  under  the  rate,  and  in  a  line  with  it 
stands  the  answer. 

Examples. 

1.  In  what  time  will  601.  per  annum,  payable  yearly,  amount  to 
£2C2-47G96,  allowing  61.  per  cent,  compound  interest,  for  (he  for- 
bearance of  payment? 

26247696=amount. 
Multiply  by  106=ratio. 

157486176 
262476960 


278-2255776 
Add     60"  =annnuity.         Or  thus ; 

Annuity— 60)262-47696=amt. 

333  2255776  *"  -- 

Subtract  262-47696  4-374616 

1,  Subtract  1* 


Divide  by  60)76-7486176 


337461G 


3. 

2-314616 
Subtract  M236     = 

=ratlo! 
=raUol 

2 

4. 

1-191016 
Subtract  1-191016= 

3^ 

rm 

4  vpars. 

Divide  by  106)1-96247696        2.  Subtract  1-06         =ratio. 

Divide  by  1  06)1-191016 

Divide  by  106)11236 

Divide  by  1-06)106 

1 

The  number  of  divisions  by    ^  Or,  fooking  into  Table  III.  un- 
106,  being  4,  gives  the  number    f '  /if;,  '!^^'/'    the   quotient 

of  years  ~  4,  the  answer.  t^'^^^^'  '^'"^^'  '»"•"''  "^  ^^^''' 

Ans.  as  before. 

Or,  in  Table  I.  under  the  given  rate,  you  will  find  1'262476,  and 

m  a  line  under  years,  you  will  find  4.     ^ 

2.  In  what  time  will  an  annuity  of  g60  payable  yearly,  amoun> 
to  §2207-1354,  allowing  6  per  cent,  for  the  forbearance  of  par 
n^ent?  Ans.  20  years/ 


320  PKESENT  WORTH  OF  ANNUITIES,  ^c. 

PRESENT  WORTH  OF  ANNUITIES,  4'Q.  AT  COMPOUND  IN 

TEREST. 

CASE  I. 

JVhen  the  annuityy  4'c.  rate  and  time  arc  given  to  find  the  present 

worth. 

Rule  I.* 
1.  Divide  the  annuity  by  the  amount  of  jjl  or  <£!  for  1  year,  and 
the  quotient  will  be  the  present  worth  of  1  year's  annuity. 

*  This  rule  depends  on  the  rule  for  finding  the  present  worth  in  Discount  at 
Compound  hiterest.  For  each  year  the  present  worth  is  to  be  found  by  that 
rule.  Then,  the  sum  of  the  present  worth  for  the  several  years,  must  evidently 
be  the  present  worth  of  the  whole,  and  is  the  rule. 

Or,  suppose  the  annuity  to  be  11.  or  $1  at  6  per  cent,  then  —  is  the  present 

l*Ot) 

■^vorth  for  one  year ;  for  two  years ; for  three  years ;  .  fov 

^        '   l-06a  ^  1-063  ^  1-06.4 

tour  years,  and  so  on.    Then  the  sum,  or 1 1 — -4 ,  will   be 

i\ic  whole  present  worth.     Let  any  annuity  be  substituted  for  the  numerator  of 
( hese  several  fractions,  and  you  have  the  rule  in  the  text. 

By  Note  2,  Prob.  I.  of  Geometrical  Progression,  the  sum  of  the  series,  

'  ^  1-U6 

-        J  1        .        "^        -    .  1  11  1  1  >x         ...., 

L ,  is  1 — -  X  — \  or X .     ^ ow  \i  W\F. 

1-063^1-063'   1-064  1-064     -06         -06        '06      1-064 

ramuity  were  to  continue  forever,  or  the  number  of  years  were  infinite,  then  the 
index  of  the  denominator  of  the  last  expression  would  be  infinite,  and  the  value 

of  the  fraction  would  be  infinitely  diminished  or  become  notliing,  and  — -  would 

-06 
be  the  present  worth  of  an  annuity  of  11. or  $1  to  continue  forever  ^t  6  per  ccnL 
Hence,  if  an  aimuity  is  is.  perpetuity^  oris  to  continue  forever,  its  present  Avorth  is 
found  by  dividing  the  annuity  by  the  ratio,  or  the  interest  of  11.  or  ^1  for  a  year 
at  the  given  rate. 

The  present  worth  of  an  annuity  of  $1  to  continue  forever  at  5  per  cent,  is 

— z==— r  =$20,  and  an  annuity  of  $100  at  5  per  cent,  to  continue  forever,  would 

now  be  worth  $2000,  and  at  7  per  cent.  $1 428A. 

Rule  II.  is  derived  from  the  expression  1 X-— ,  when  the  annuity  is 

1-064  'Ob 
^1  or  11.  and  the  rate  6  percent.  That  is  wlien  the  annuity  is  $1  or  11.  divide 
the  annuity  by  that  power  of  the  ratio  indicated  by  the  number  of  years,  and 
subtract  the  quotient  from  the  annuity  ;  the  remainder  divided  by  the  ratio  of  the 
series  less  1,  will  be  the  present  worth.  But  the  present  worth  of  annuities  va- 
ries as  the  annuity.     Hence  the  rule  is  manifest. 

Note.     Another  rule  for  obtaining  the  present  worth  may  be  derived  from  the 

preceding.     Thus,  the  sum  of  the  series, -J-,  —-.  ~,  —Z  is,  by  Note 

2,  of  Prob.  I,  of  Geometrical  ProgrGssion,  1 X ,  which    is  alro 

**  1.064      1-06—1 

X •= — : =rtbe  present  worth  of  11.  or  gl   for  four 

1-4)64         l-ed--l     1-06*— 1064  ^  ' 

year^  at  6  per  cent.     That  i?.  divide  the  difference  between  unity  and  that  power 


AT  COMPOUND  INTEREST.  321 

2.  Divide  the  annuity  by  the  square  of  the  ratio,  anil  the  quo- 
tient will  be  the  present  worth  for  two  years. 

3,  In  like  manner,  find  the  present  worth  of  each  year  by  itself, 
and  the  sura  of  all  these  will  be  the  present  value  of  the  annuity 
sought. 

Rule  II. 
Or,  divide  the  annuity,  &lc.  by  that  power  of  the  ratio  signified 
by  the  number  of  years,  and  subtract  the  quotient  from  the  annui- 
ty ;  this  remainder  being  divided  by  the  ratio  less  1,  the  quotient 
will  be  the  present  worth. 

ElXAMPLES. 

1.*  What  ready  money  will  purchase  an  annuity  of  601.  to  con- 
tinue 4  years,  at  61.  per  cent,  compound  interest  ? 

First  Method. 
Ratio    =  106)60-00000(56-603=present  worth  for  1  year. 

Kaibl^=r         11236)6000000(53-399=         do.  for  2  years. 

Katiol''=     M91O16)i30OO000(50-377=         do.  for  3  years. 

Kaliol^  =l-26247696)60-00000(47-525=:         do.  for  4  years. 

207-904=i:2O7   13s.  0|d.  Ans. 


be^raUo^^^  =l-26247696)6O0000000(47-525 


Second  Method 

4th 

the 

From  60  ^^ 

Subtract  47-625  Or,  4--^ =47-525  GO— 47-525=12-475 

-Odl  12'475 

Divis.l-OG— 1=-06)12-475  And =r:207-916- 

-06 

2  07-916=je207  18s.  3^-d.  Ans. 

of  the  ratio  which  is  indica'ted  by  the  number  of  years,  by  the  difference  between 
that  power  of  the  ratio  which  is  one  greater  than  the  number  of  years  and  that 
power  of  the  ratio  which  is  equal  to  the  number  of  years,  and  the  quotient  is  tlie 
present  worth  of  11.  or  ^1.  Then,  as  annuities  are  as  their  present  worth,  multi- 
ply this  quotient  by  the  given  annuity,  and  the  product  is  its  present  worth.  The 
rules  for  the  next  Case  are  derived  directly  from  this  rule,  and  need  no  further 
illustration. 

*  The  amount  of  an  annuity  may  also  be  found  for  years  and  parts  of  a  year, 
thus  : 

1.  Find  the  amount  for  the  whole  years,  as  before. 

2.  Find  the  interest  of  that  amount  for  the  given  parts  of  a  year. 

3.  Add  this  interest  to  the  former  account,  and  it  will  give  the  whole  amount 
required. 

The  present  worth  of  an  annuity  for  years  and  parts  of  a  year  may  be  found 
thus : 

1.  Find  the  present  worth  for  the  whole  years,  as  before. 

2.  Find  the  present  worth  of  this  pi-esent  worth,  discounting  for  the  given 
parts  of  a  ystr,  and  it  will  be  tlie  whole  present  worth  required. 

■Questions  in  this  case  may  also  be  answered  by  first  finding  the  amount  of 
ihG  given  annuity  by  Case  I.  of  annuities  in  arrears,  page  315,  and  then  the 
present  worth,  or  principal,  by  Case  II.  of  Compound  Interest, -page  311. 

R  r 


322 _  PRESENT  WORTH  OF  ANNUITIES,  Lc. 

By  Table  III. 

Under  61.  per  cent,  and  opposite  4,  we  find 

4-3746  l=amount  of  11.  annuity  for  4  year?. 
Multiply  by  60=annuity. 

262-47660=amount  of  601.  for  4  years. 
Then,  opposite  4  years,  and  under  61.  per  cent,  in  Table  2(1. 
We  have  -792093 
Multiply  by  262-7466 

4762558 
4752568 
3168372 
5544651 
1584186 
4752558 
1584186 


208.1 197426338=  £208  2s.  4id. 
Or,  opposite  4  years,  and  under  61.  per  cent,  in  Table  1st,  we 
have  l-26247=the  anjount  of  11.  for  4  years  : 

Then,  262-7466-t-I -26247=208- 1209=  £208  2s.  5d.  Ans. 

By  Table  IV.* 
Multiply  the  tabular  nuKiber,  under  the  rate,  and   opposite  the 
time,  into  the  annuity,  and  the  product  will  be  the  present  worth. 
Thus,  in  Example  1st.    What  ready  money  will  purchase  £60 
annuity,  to  continue  4  years,  at  £6  per  cent,  compound  interest  ? 
Under  61.  per  cent,  and  even  with  4  years. 

We  have  3-465 l=present  worth  £l  for  4  year?. 
Multiply  by         60=annuity. 

Ans.=207-9060=  £207  18s.  lid. 
2.  What  is  the  present  worth  of  an  annuity  of  ^60  per  annum, 
to  continue  20  years,  at  6  per  cent,  compound  interest  ? 

Ans.  $688-65  (nearly.) 

CASE  II. 
When  the  present  tvorth,  time^  and  rate  are  given)  to  find  the  annuity', 

rent,  4*c. 

Rule. 
1.   From  that  power  of  the  ratio,  denoted  by  the  number  of 
years  plus  I,  subtract  that  power  of  it  denoted  by  the  number  pf 
years. 

*  Table  4th  is  thus  made  :  Divide  £1  by  l-06='94339  the  present  worth  of 
the  first  year,  which,  divided  by  1-06,  is  equal  to  -88999,  which,  added  to  the  first 
year's  present  wortli,  is  =  1-83339,  the  second  year's  present  worth,  then -88999, 
divided  by  1-06,  and  the  quotient  added  to  1-83339,  gives  2-6701  for  the  tliirU 
year's  present  worth,  &c. 


i 


AT  COMPOUND  INTEREST.  323 

2.  Divide  the  remainder  by  that  power  of  the  ratio,  signified  by 
the  time  made  less  by  unity. 

3.  Multiply  the  present  worth  into  this  quotient,  and  the  product 
will  be  the  annuity,  pension,  rent,  &c. 

Or,  1.  Multiply  that  power  of  the  ratio,  denoted  by  the  number 
of  years  plus  1,  by  \he  present  worth. 

4.  Multiply  that  power  of  the  ratio,  denoted  by  the  time,  by  the 
present  worth,  and  subtract  this  product  from  the  former. 

5.  Divide  the  remainder  by  that  power  of  the  ratio,  denoted  by 
the  time  made  less  by  unity,  and  the  quotient  will  be  the  annuity. 

Examples. 
1.  What  annuity,  to  continue  4  years,  will  £207*904  purchase, 
compound  interest,  at  £6  per  cent.  ? 

First  Method. 
From  l-06xl-06xl-06xl'06xl-06=l-3382255776 
Subt.   l-06xl-06xl-06xl-06  =1-26247696 


Divide  by  1-06|4— 1=-26247696)'0767486176('2885898 
•2885898 
Multiply  by       207-9  present  worth. 

25973082 
20201286 
57717960 


Ans.  59-99781942=  £60. 

Second  Method. 
From  1-06X1-06XI-06X1-06X  t-06x207-9=278-21709757;3 
Take  1-Oex  1*06  X 1-06  Xl'06x207-9  =262-468959984 


Divide  by  1-06|*  — 1  =-26247696)15-748137589(59'998=601, 
By  Table  V.* 

Multiply  the  tabular  number  corresponding  with  the  rate  and 
time,  by  the  purchase  money,  and  the  product  will  be  the  annuity. 
Under  £6  per  cent,  and  opposite  4  years,  you  will  find 

•28859=annuity  which  £l  will  purchase  in  4  years. 
Multiply  by     207-9 

259731 
202013 
577180 


59-997861  =  £60. 
2.  What  salary,  to  continue  20  years,  will  g688  66c.  purchase, 
at  6  per  cent,  compound  interest  ?  Ans.  ^60. 

*  Table  5th  is  made  in  this  manner :  Divide  £1  by  the  present  worth  of  jgl 
for  1  year,  and  the  quotient  will  be  the  annuity,  which  £l  will  purchase  for  1 
year  :  divide  £l  by  the  present  worth  of  £l  for  2  years,  and  the  quotient  will 
be  the  annuity,  which  Jgl  will  purchase  for  2  years,  &c. 


224  ANNUITIES,  kc.  IN  REVERSION 

CASE  III. 

When  the  annuity  ^  present  worth  and  ratiOy  are  given^  to  find  the  time. 

Rule.* 
Divide  the  annuity  by  the  product  of  the  present  worth  and  ra- 
tio subtracted  from  the  sum  of  the  present  worth  and  annuity,  and 
the  quotient  will  be  that  power  of  the  ratio,  denoted  by  the  num- 
ber of  yearg,  which,  being  divided  by  the  ratio,  and  this  quotient 
by  the  same,  till  nothing  remain,  the  number  of  divisions  will  show 
the  time  :  Or,  the  above  quotient  being  sought  in  Table  1st  under 
the  given  rate,  in  a  hue  with  it,  you  will  see  the  time. 

Examples. 
1.  For  bow  long  may  an  annuity  of  £60  per  annum  be  purchas- 
ed for £207-906336762",  at £6  per  cent,  compound  interest  ? 
Multiply  207-906336762  To  207-906336762==present  worth, 

by  1-06  Add    60-  =annuity. 

1247438020572         From  267-906336762 
2079063367620  Subt.  220-380716967 


220-38071696772  47-525619793=divisov. 

47  6256 1 9795)60-000000000(  1  '26247696 
Divide  by  1-06)1-26247696 

1  06)1-191016 

1-06)1  •1236-. 

1-06)  1-06 

J  }  The  number  of  divisions 
P^     '  =time=4  years. 

Or ■         —  =  1-26247696, 

'207-906336762-f60— 207-906336762x1 -06 
which  heing  sought  in  Table  1,  under  the  given  rate,  in  a  line  with 
it,  is  4=4  years. 

2.  How  long  may  a  lease  of  ^300  yearly  rent,  be  had  for 
«s2132-341  allowing  5  per  cent,  compound  interest,  to  the  purchas- 
er? Ans,  9  years. 

AKHUITIES,  LEASES,  4-c.  TAKEjY  I/Y  REVERSI0,X  AT  COM- 
POUND  INTEREST. 

CASE  I. 

When  the  annvilij,  time  and  ratio,  are  gi;&en^  to  find  the  present  Torth 
of  the  annuity  in  reversion. 

RULF,    I. 

1.  Divide  the  annuity  by  that  power  of  the  ratio  denoted  by  the 
tir?te  of  its  continuance. 

■  Tius  rale  lo  derived  directly  from  Rule  II.  Case  L 


I 


AT  COMPOUND  INTEREST.  325 

2.  Subtract  this  quotient  from  tlie  annuity:  divide  by  the  ratio 
less  1,  and  the  quotient  will  be  the  present  worth,  to  commence 
immediately. 

3.  Divide  this  quotient  by  that  power  of  the  ratio  denoted  by 
the  time  of  reversion,  (or,  time  to  come,  before  the  annuity  com- 
mences) and  the  quotient  will  be  the  present  worth  of  the  annuity 
in  reversion. 

EULE  II. 

Or,  1.  Multiply  the  annuity  by  that  power  of  the  ratio  denoted 
by  the  time  of  its  continuance,  minus  unity,  for  a  dividend. 

2.  Multiply  that  power  of  the  ratio  denoted  by  the  time  of  its 
continuance,  that  power  of  it  denoted  by  the  time  of  reversion, 
and  the  ratio  less  1,  continually  together  for  a  divisor,  and  the  quo- 
tient arising  from  the  division  of  these  two  numbers  will  be  the 
present  worth  of  the  annuity  in  reversion. 

Rule  III. 

Find  the  present  worth  of  the  annuity  for  the  number  cf  year- 
before  it  is  to  begin  and  is  to  be  continued  ;  find  also  the  present 
worth  of  the  annuity  for  ihe  number  of  years  before  the  annuit  v 
commences :  and  the  difterence  between  the  present  worth  for 
these  two  periods,  is  the  present  worth  of  the  annuity  in  rever- 
sion.* 

Examples. 

1.  What  is  the  present  worth  of  601.  payable  yearly,  for  4  years  ; 
Tiut  not  to  commence  till  two  years  hence,  at  61.  per  cent.  ? 

First  Method. 

Ratio=l-06       Or,  in  Table  4th,  find  the  prescriL 
1'06  value  of  11.  at  the  given  rate,  both  for 
the  time  in  being  and  the  time  in  re- 


636  version  added  togetlier.  and  subtract 
1060     the  present  worth  of  the  ti.ne  in  be- 
ing from  the  other,  multiply  the  re- 


2d.  power=l'1236  mainder  by  the  annuity,  and  the  pro 
Carried  over.  11236  duct  will  be  the  answer. 


*  Rule  III.  is  merely  an  expression  of  this  truth,  viz.  the  present  worth  of  au 
annuity  continued  for  the  sum  of  the'  years  before  the  annuity  is  to  commence, 
and  is  to  continue  after  it  begins,  is  evidently  too  much  by  the  present  worth  of 
the  same  annuity  for  the  time  of  reversion  or  the  time  before  the  annuity  is  to 
commence, 

Rvue  I.  The  first  two  steps  are  Rule  II.  of  Case  I.  to  find  the  present  worth 
of  Annuities  &lc.  at  Compound  Interest,  for  iJio  time  of  reversion  and  continu- 
ance of  the 'annuity.  '  But  as  this  present  worth"  would  evidently  bs  Ico  much, 
it  must  be  discounted  at  compouud  interest  for  the  time  of  reversion,  which  in 
the  f  rst  example  is  2  years.  The  third  step  in  R,ul3  I.  is,  tlierefore,  meroly  tlie 
rule  for  Discount  at  Compound  Interest.     Hence  the  process  is  obvious. 


''W^^i^  -  4HBp 


w 


o2G  ANNUITIES,  &c.  IN  REVERSION 

Brought  OFer.        1-1236  Pres.  worth  of  the  time 


in  being  and  reversion  5        '^^  ^^"^ 

64416  Present  worth   of  the  i  _^  ^^ 

33708  time  in  being                )  ""ilirri. 

22472  308402 

11236  60 

11236  

£18504120 


Div,  by  4th  pow.=l-26247696)60'O00000000000(47'525619794281 
Subtract  the  quotient=47-625619794281 

Divide  by  1'06— 1='06)12-474380206719 

Divide  by  l-06xl-06=M236;)207-9063367619(185'035899=185l. 
Os.  8id.=the  present  worth  of  the  annuity  in  reversion. 
60  60—47-5256 

^^'  ri62i^96=47-5256         -^^-^=207-906 

207-906 
And    J. j^^g  =185-035899 

Second  Method. 
•26247696=4th  power— 1 
Multiply  by  60=annuity. 

15-74861760=dividend,      •08511115)15-74861760(185-03a 
l-26247696=4th  power.  [Ans. 

M236=2d  power. 

757486176 
378743088  1-06|4— 1x60 

.252495392         Or,  "7:7^^,4^-7:7^13^.^^-^=125 -036 
126247696  ^^^1  X^^^'  Xl*06— 1 

126247696 


1-418519112256 

•06=ratio— 1 


•0851 1 1 14673536=divisor. 

Third  Method. 


11 
60x1—  ■f:56lX:5^=295038-^  the  present  worth  for- 6  years. 

60x1— -j^^X.-^=  11 0-002 -}-  the  present  worth  for  2  years. 


£l85-035-f =the  present  worth  in  reversion. 

Example  2. 
What  is  the  present  worth  of  a  reversion  of  a  lease  of  ^60  per 
annum,  to  continue  20  years,  but  not  to  commence  till  the  end  of  8 
\earsj  allowing  6  per  cent  to  the  purchaser? 

Ans.  ^431-782  (nearly.; 


AT  COMPOUND  INTEREST.  327 

3.  An  annuity  of  §1  in  reversion  is  to  commence  at  the  end  of  20 
years,  and  is  to  continue  15  years  ;  what  is  its  present  worth  at  4 
percent.?  '  Ans.  §60743  nearly, 

4.  An  annuity  of  gl  in  reversion  is  to  commence  after  5  years, 
and  to  continue  forever  ;  what  is  its  present  worth  at  6  per  cent.  ? 

Ans.  gl2  45c.  43. 

w3rt  annuity i  several  times  in  reversion^  and  rate  being  given,  to  find 
the  several  present  values. 

Find  the  present  value  of£l  6rf,l  by  Table  4,  at  the  given  rate, 
and  for  the  several  given  times,  which,  being  severally  multiplied 
by  the  annuity,  the  products  will  be  the  several  present  values  of 
that  annuity,  for  the  several  times  given  ;  subtract  the  several 
present  values,  the  one  from  the  other,  and  the  several  remainders 
will  answer  the  question. 

6.  A  has  a  term  of  6  years  in  an  estate  at  601.  per  annum.  B 
has  a  term  of  14  years  in  the  same  estate,  iu  reversion,  after  the  6 
years  are  expired  ;  and  C  has  a  further  term  of  16  years,  after  the 
expiration  of  20  years.  I  demand  the  present  values  of  the  seve- 
ral terms  at  6  per  cent.  ? 

£  s.     d. 
Pres.  value  of  £1  for  36y.==14-61722x60==877  0     7-2- 
Ditto  of  ditto  for  20  years  =11-46992x60=688  3  10^- 
Ditto  of  ditto  for  6  years    =  4-91732x60=295  0     9a=A's  term. 
Therefore,    877  0  7f--688  3  10|=je  188  16  9    C's   term,  and 
688  3  10^—295  0  9i=£393  3  U=B>8  term. 

6.  For  a  lease  of  certain  profits  for  7  years,  A  offers  to  pay  ^300 
gratuity,  and  §300  per  annum,  B  offers  §800  gratuity  and  §250  per 
annum,  C  bids  §1300  gratuity  and  §200  per  annum,  and  D  bids 
§2500  for  the  whole  purchase,  without  any  yearly  rent  ;  which  is 
the  best  offer,  computing  at  6  per  cent.  ?  § 

By  Table  4,  the  present  worth  of  §300  per  annum  }   -^-.  «, . 
for  7  years,  at  6  per  cent  is  S   ^*"'**'  ^ 

To  which  add     300* 


Value  of  A's  offer=1974-714 


Present  worth  of  §250  per  annum  for"  7  years=1395-595 

To  which  add       800- 


Value  of  B's  offer=2195-595 


Present  worth  of  §200  per  annum  for  7  years=ll  16-476 

To  which  add     1300- 


Value  of  C's  offer=24 16-476 


D's  offer=2500- 
Hence  it  appears  that  D-s  offer  is  the  best. 
The  above  questions  may  be  answered  by  the  4ih  arid  2d  Tables. 


3i5  ANNUITIES,  &c.  IN  REVERSION 

Take  question  Isi.for  Example. 

1.  Blultiply  the  tabular  number  in  Table  4,corricsponding  to  the 
rate  and  the  time  of  continuance,  into  the  annuity,  and  the  pro- 
duct will  t>e  the  present  worth,  to  commence  immediately. 

2.  Multiply  this  present  worth  by  the  tabular  number  in  Table  2, 
corresponding  to  the  rate  and  the  time  of  reversion,  and  the  pro- 
duct will  be  the  present  worth  of  the  annuity  ia  reversion. 

In  Table  4th  we  have  3-4651 

Multiply  by  60=annuity. 

207-90G0 
in  Table  2d  we  have  -esg&DG 


1247436 
1871134 
1871154 
1871154 
•  1663248 

1663248 

185  035508376=prc3.  worth  of  the  reversion. 

CASE  11. 

When  {he  present  worth  of  the  reversion y  rate  and  time  are  giverij  to 
find  the  annuity. 
Rule  1.  Multiply  that  power  of  the  ratio  signified  by  the  time 
of  reversion,  by  the  present  worth,  and  the  product  will  be  the 
amount  of  the  present  worth  for  the  time  before  the  annuity  com- 
mences. 

2.  Multiply  that  power  of  the  ratio  signified  by  the  time  of  con- 
tinuance plus  1,  by  the  last  product. 

3.  Multiply  that  power  of  the  ratio,  signified  by  the  time,  by  the 
aforesaid  product,  and  this  last  product,  divided  by  that  power  of 
the  ratio  denoted  by  the  time  minus  unity,  will  give  the  annuity. 

Or,  Divide  the  continual  product  of  the  present  worth,  ihat 
power  of  the  ratio  denoted  by  the  time  of  continuance,  that  j)ower 
of  it  denoted  by  the  time  of  reversion,  and  the  ratio  minus  1,  by 
that  power  of  the  ratio  denoted  by  the  tinie  of  continuance  minus  J, 
and  the  quotient  will  be  the  annuity. 

Examples. 
1.  What  annuity,  to  be  entered  upon  2  years  hence,  and  then  to 
continue  4  years,  may  be  purchased  for  gl85  03^899,  at  6  f>er  ct.  ? 
First  Method. 
1  06x1  06=  l-l236==2d. power  ofthe  ratio. 
Multiply  by  185  Q36=pre*ent  wortii. 

,     67416 
■     33708 
5.61800 
89888 
11236 


207-9064496  amount  for  the  lime  of  reversion. 


AT  COMPOUND  INTEREST.  329 

3rou^ht  up.         207-9064496  amount  for  the  time  of  reversion. 
Multiply  by  I  33822  =:3th  power  of  the  ratio. 

415812  4th  power  of  the  ratio= 1-26247 

415812  Multiply  by  207-906 

1663248  

623718  757482 
623718  11362230 
207906  883729 
2524940 


From  278-22396732 


Take  262-47508782  26247508782 


Divide  by  l-06|4--l  =  -26247)15-7488750(60  the  annuity  required. 
Or,  185-036xM236=207-906 


207  •906X 1  -33822— 207-  906x  1  -26247 
Then, 1  •26247— 1 "^^^^  ^*"^- 

Second  Method. 

Jl85036=present  worth  of  the  reversion. 
l-26247=4th  power  of  the  ratio. 

1295252       Or  by  Table  4th,  divide  the 

740144     present  worth  of  the  reversion 

370072       by  the  difference  between  the 

1110216         present  worth  of  j^l  for  the  time 

370072  both  in  being  and  reversion,  and 

185036  the  time  in  being,  and  the  quo- 

tient  will  be  the  annuity. 

233-6024 

l-1236=2d.  power  of  the  ratio. 


1401614-1  oi7'^9—  5  P**'  ^^^^^  ^H^  ^^  ^^^ 
7008072  ^'^J^^^—  ^  time  in  being  k  revsn. 
4672048  _.  ^  present  worth  of  gl  for 


2336024  1  ^^^-^^  —  ^      the  time  in  being 

2336024  

3'O8402)185'04!2(60  Ans. 


262-47565664 

•06=ratio— 1. 


-;.-0tJi4_i=:.o6247)15  7485393984(60.  

185-036x1 -26247X1  •1236x1-06—1 

^^'  r26247i:i  =='^^' 

2.  The  present  worth  of  a  lease  of  a  house  is  iG431  15s.  7d. 
2  7819qrs.  taken  in  reversion  for  20  years  ;  but  not  to  commence 
till  the  end  of  8  years,  allowing  £6  per  cent,  to  the  purchaser : 
What  is  the  yearly  rent?  Anj5.  £60. 

^   s 


<5aU  PURCHASING  FREEHOLD  ESTATES 

PURCmSljVG  AJVJVUITIES  FOREVER,  OR  FREEHOLD  ES 
TAXES,  AT  COMPOUND  IJVTEREST. 

CASE  I. 

JVhen  the  annuity^  or  yearly  rent,  and  the  rate  are  giveUf  to  find  ike 
present  worth  or  price. 

Rule.* 
As  the  rate  per  cent,  is  to £100  or  $100  so  is  the  yearly  rent,  to 
the  Talue  required. 

Or,  Divide  the  yearly  rent  by  the  ratio  less  1,  and  the  quotient 
will  be  the  value  required. 

Examples. 
1.  What  is  the  worth  of  a  freehold  estate  of  £60  pel:  annum,  al- 
lowing 61.  per  cent,  to  the  purchaser  ? 
£      £        £ 
6  :  100  ::  CO  Or,  I'OG— 1=:'06)60'00 

60  ■ 

- —  1000 

6)6000 


•     •       £1000  Ans. 

2.  An  estate  brings  in  yearly  $15  :  What  will  it  sell  for,  allow- 
ing the  purchaser  6  per  cent,  compound  interest?    Ans.  ^1500. 

CASE  n. 

When  the  price,  or  present  worth,  and  rate  are  given,  to  find  the  an- 
nuity y  or  yearly  rent. 

Rule. 
As£100  or  glOOis  to  the  rate  go  is  the  present  worth  to  its  rent. 
Or,  Multiply  the  present  worth  by  the  ratio  less  1,  and  the  pro- 
duct will  be  the  yearly  rent. 

Examples. 

1.  If  a  freehold  estate  be  bought  for  £1000  allowing  £6  per 
cent,  to  the  purchaser  :   What  is  the  yearly  rent  ? 

£     £        £ 
400  :  6  ::  1000 
6 

Or,  1000  X  -06=  £60. 

100)6000(£60  Ans. 
600 

0 

2.  If  an  estate  be  sold  for  gl500  and  5  per  cent,  allowed  (o  (he 
buyer  ;  what  is  the  yearly  rent  ?  Al)s.  $15. 

*  The  reason  of  this  rule  is  obvious  ;  for  since  a  year's  interest  of  the  prico» 
•which  is  given  for  it,  is  the  annuity,  there  can  neither  more  nor  less  be  made  oi 
that  price,  than  of  the  aimuity,  whether  it  be  employed  at  simple  or  compound 
interest.  It  has  also  been  proved  under  Case  I.  of  the  Present  Worth  of  Annu- 
ities &:c.  at  Compoond  Interest.  Case  II.  and  HI.  follow  dirc6tly  from  the  rule 
for  Case  I.  and  their  rules  are  hence  manifest. 


AT  COMPOUND  INTEREST.  33i 

CASE  HI. 

When  the  present  worthy  or  price,  and  yearly  rent,  are  given,  to  find 

ike  rale. 

Rule. 

As  the  present  worth  is  to  the  rent ;  so  is  £100  or  ^100  to  the 
u'ate. 

Or,  Divide  the  rent  by  the  present  worth  ;  add  1  to  the  quo- 
tient, and  the  sum  will  be  the  ratio  of  the  rate  per  cent. 

Or,  Divide  the  sum  of  the  present  worth  and  rent  by  the  pres- 
ent worth,  and  the  quotient  will  be  the  ratio. 

Examples. 

1.  If  an  estate  of  £60  per  annum  be  bought  for  £1000  what 
rate  of  interest  was  allowed  the  purchaser  for  his  money  ? 
£        £        £ 

1000  :  60  ::  100  Or,  1 000)60 •00(-06-l-l==l '06 

100  6000 


1000)6000(£6  Ans. 


Or,  to  1000=present  worth. 
Add      60=rent. 

1000)1060(1-06 
1000 


6000 
6000 
2.  An  estate  of  §75  per  annum  was  purchased  for  ^1500  what 
rate  of  interest  bad  the  buyer  for  his  money  ?     Ans,  5  per  cent. 

To  find  at  how  many  years'  purchase  an  estate  may  be  bought. 

CASE  I. 

JVhen  the  rate  of  interest  is  given,  to  find  the  number  of  years. 

Rule. 

Divide  £100  or  g  100  by  the  rate,  and  the  quotient  will  be  the 
|v  years. 

Examples. 

1.  How  many  years'  purchase  should  a  gentleman  offer  for  the 
I  purchase  of  an  estate,  to  have  6  per  cent,  for  his  money  ? 

6)100 

16  666-f  =  16|  years. 

2.  How  many  years'  purchase  is  ah  estate  worth,  allowing  5  per 
;ent.  to  the  purchaser  ?  Ans.  20  years^ 


332  PURCHASING  FREEHOLD  ESTATfcs 

CASE  n. 

When  the  number  of  years*  purchase,  at  -which  an  estate  is  bought.  Or 
sold,  is  given,  to  find  the  rate  of  interest. 

Rule. 

Divide  £lOO  or  ^100  by  the  number  of  years,  and  the  quotient 
will  be  the  rate. 

Examples. 

1.  A  gentleman  gives  16|  years'  purchase  for  a  farm  ;  what  in- 
terest is  he  allowed  ?         16|=16-666-|-)100000(6  per  cent.  Ans, 

2.  A  gentleman  gives  20  years'  purchase  for  an  estate  ;  what  in- 
terest has  he  ?  Ans.  5  per  cent. 

PURCHASIJVG  FREEHOLD  ESTJITES  IJV  REVERSION. 

CASE  I. 

The  rate  and  rent  of  a  freehold  estate  being  given,  to  find  the  present 
worth  of  reversion. 

Rule.* 

1.  Find  the  present  worth  of  the  annuity  or  rent,  (by  Case  1, 
of  purchasing  Freehold  Estates,  page  330,)  as  though  it  were  to 
be  entered  on  immediately. 

2.  Divide  the  last  present  worth  by  that  power  of  the  ratio  dct 
noted  by  the  time  of  reversion  (by  Discount  by  Compound  Inter- 
est) and  the  quotient  will  be  the  answer  required. 

Or,  1.  Having  found  the  present  value  of  the  estate,  supposing  it 
to  be  immediate  :  Multiply  the  annuity,  or  rent,  by  the  present 
worth  of  II.  or  $1  corresponding  with  the  time  of  reversion  and 
rate  in  Table  4th,  and  the  product  will  be  the  present  worth  of 
the  annuity,  or  rent,  for  the  time  of  reversion  ;  or  the  value  of  the 
present  possession. 

2.  Subtract  the  value  of  the  possession  from  the  value  of  the 
estate,  and  the  remainder  will  be  the  value  of  reversion. 

Examples. 

1.  Suppose  a  freehold  estate  of  601.  per  annum  to  commence  2 
years  hence,  be  put  up  to  sale  ;  what  is  its  value,  allowing  the 
purchaser  61.  per  cent.  ? 

Fii^st  Method. 
l-Oe— l='06)60'00=rent  per  annum. 

1000=present  worth,  if  entered  on  immediately. 

*  By  the  first  step,  the  present  worth  is  found  for  the  present  time ;  but  as  the 
estate  is  not  to  be  entered  on  for  a  certain  time,  discount  for  that  time  must  be 
allowed  at  Compound  Interest.     This  is  the  second  step,  and  the  jDropriety  of  j 
the  rule  is  manifest.    Case  11.  needs  no  illustration. 


IN  REVERSION,  533 

F06|^  =  l-1236)1000'000(889996=£889  19s.  lid.— present  worlh 
of  10001.  for  2  years,  or  the  whole  present  worth  required. 

Second  Method. 
1.06—1= •06)60-00 

1000=present  worth,  for  immediate  possession. 
In  Table  4th.  we  have  l-33339=value  of  II.  for  2  years. 
Multiply  by  60=rent. 

110'00340=value  of  possession. 
From        1000-OpOO 
Subtract  110-0034 


889*9966=value  required. 
2.  Suppose  an  estate  of  §75  per  annum,  to  commence  10  years 
hence,  were  to  be  sold,  allowing  the  purchaser  5  per  cent ;  what 
is  its  worth  ?  Ans.  $920  87c.  Im.  (nearly.) 

CASE  II. 

The  value  of  a  Reversion y  the  Time  prior  to  its  Commencemeyit,  and 
rate  of  Interest  given,  to  find  the  Annuity  or  Rent. 

Rule. 

1.  Multiply  the  price  of  the  reversion  by  that  power  of  the 
amount  of  11.  or  gl  for  1  year,  denoted  by  the  time  of  reversion, 
and  the  product  will  be  its  amount,  (by  Case  1  of  Compound  In- 
terest.) 

2i  Find  the  interest  of  the  amount  (by  Case  1st  Simple  Interest) 
and  it  will  be  the  annuity,  or  yearly  rent. 

Examples. 
1.   A  freehold  estate  is  bought  for  £889*9966  which  does  not 
commence  till  the  end  of  2  years  ;  the  buyer  being    allowed  €1. 
per  cent,  for  his  money  ;  1  desire  to  know  the  yearly  income  ? 
889-9966=price  of  the  reversion. 

Multiply  by  Toel^  =  1-1236  denoted  by  the  time  of  reveriioH. 

63399796 
26699898 
17799932 
8899966 
8899966 


1000'00017976=amount  of  the  reversioa. 

•06 


Ans.  £60-00 
2.  If  a  freehold  estate,  to  commence  10  years  hence,  be  sold  iov 
^920  87c.   Im.   allowing  the  purchaser  T)  per  cent.  ;  what  »<?  the 
yearly  income  ?  '      Ans.  p5. 


TABLES. 


TABLE  I. 


SHEWING  THE  AMOUIVT  OF  £l    OR  $1    FROM  1 

YEAR  TO  50. 

years. 

3  per  cent. 

3i  per  cent. 

4  per  cent. 

4^  per  cent. 

1 

1-0300000 

1-0350000 

1-0400000 

1-0450000 

2 

1-0609000 

1-0712250 

1-0816000 

1-0920250 

3 

1-0927270 

1-1087178 

1-1248640 

1-1411661 

4 

1-1255088 

1-1475230 

1-1698585 

1-1925186 

5 

1-1592740 

1-1876863 

1-2166529 

1-2461819 

6 

1-1940523 

1-2292553 

1-2653190 

1-3022601 

7 

1-2298738 

1-2722792 

1-3159317 

1-3608618 

8 

1-2667700  j 

1-3168090 

1-3685690 

1-4221006 

9 

1-3047731 

1-3628973 

1-4233118 

1-4860951 

10 

1-3439163 

1-4105987 

.1-4802842 

1-5529694 

11 

1-3842338 

1-4599697 

1-5394540 

1-6228530 

12 

1-4257608 

1-5110686 

1-6010322 

1-6958814 

13 

1-4685337 

1-5639560 

1-6650735 

1-7721961 

14 

1-5125897 

1-6185945 

1-7316764 

1-8519449 

15 

1-5579674 

1-6753488 

1-8009435 

1-9352824 

16 

1-6047064 

1-733986 

1-8729812 

2-0223701 

17 

1-6528476 

1-7946755 

1-9479005 

2-1133768 

18 

1-7024330 

1-8574892 

2-0258161 

2-2084787 

19 

1-7535060 

1-9225013 

2-1068491 

2-3078603 

20 

1-8061112 

1-9897888 

2-1911231 

2-4117140 

21 

1-8602945 

2-0594314 

2-2787680 

2-5202411 

22  ■ 

1-9161034 

2-1315115 

2-3699187 

2-6336520 

23 

1-9735865 

2-2061144 

2-4647155 

2-7521663 

24 

2-0327941 

2-2833284 

2-5633041 

2-8760138 

25 

2-0937779 

2-3632449 

2-6658363 

3-0054344 

26 

2-1565912 

2-4459585 

2-7724697 

3-1406790 

27 

2-2212890 

2-5315671 

2-8833685 

•  3-2820095 

28 

2-2879276 

2-6201719 

2-9987033 

3-4296999 

29 

2-3565655 

2-7118779 

3-1186514 

3-5840364 

30 

-  2-4272624 

2-8067937 

3-2433975 

3-7453181 

31 

2-5000803 

2-9050314 

3-3731334 

3-9138574 

32 

2-5750827 

3-0067075 

3-5080587 

4-0899810 

33 

2-6523352 

3-1119423 

3-8483811 

4-2740301 

34 

£-7319053 

3-2208603 

3-7943163 

4-4663615 

35 

2-8138624 

3-3335904 

3-9460889 

4-6673478 

36 

2-8982783 

3-4502661 

4-1039325 

4-8773784 

37 

2-9852260 

3-5710254 

4-2680898 

5-0968604 

38 

■  3-0747834 

3-6960113 

4-4388134 

5-3262192 

39 

3-1670269 

3-8253717 

4-6163659 

5-5658990 

40 

3-2620377 

3-9592597 

4-8010206 

5-81646^15 

41 

3-3598988 

4-0975337 

4-9930614 

6-0782054 

42 

3-4606958 

4-2412579 

5-1927833 

6-3517246 

43 

3-5645167 

4-3897020 

5-4004952 

6-6375522 

44 

3-6714522 

4-5433-115 

5-616515 

6-9362421 

45 

3-7815957 

4-7023585 

5-8411756 

7-248373 

46 

3-895U436 

4-8G69411 

6-0748236 

7-5745497 

47' 

4-0118949 

5-0372840 

6-3168166 

7-9154045 

48 

4-1322518 

5-2135889 

6-5694892 

8-2715977 

49 

4-2562193 

5-3960645 

6-8322688 

8-6438196 

50 

4-3830059 

5-5849268 

7-1055596 

9-0327915 

TABLES. 
TABLE  L— CONTINUED. 


33; 


i 


SHEWING  THE  AMOUNT  OF  £l   OR 

$1    FR03I  1  YEAR,  TO  50. 

years. 
1 

5  per  cent. 

5i  per  cent. 

6  per  cent. 

7  per  cent. 

1-0500000 

1-0550000 

1-0600000 

1-07000 

2 

1- 1025000 

1-1130250 

1-1236000 

1-14490 

3 

1-1576250 

1-1742413 

1-1910160 

1-22504 

4 

1-2155062 

1-2388245 

1-2624796 

1-31079 

5 

1-2762815 

1-3069598 

1-3382256 

1-40255 

6 

1-3400956 

1-3788426 

1-4185191 

1-50073 

7 

1-4071004 

1-4546789 

1-5036302 

1-60578 

8 

1-4774554 

1-5346862 

1-5938480 

1-71818 

9 

1-5513282 

1-6190939 

1-6894789 

1-83845 

10 

1-6288946 

1-7081440 

1-7908476 

1-96715 

11 

1-7103393 

1-8020919 

1-8982985 

2-10485 

12 

1-7958563 

1-9012069 

2-0121964 

2-25219 

13 

1-8856491 

2-0057732 

2-1329282 

2-40984 

14 

1-9799316 

2-1160907 

2-2609039 

2-57853 

15 

2-0789281 

2-2324756 

2-3965581 

2-75903 

16 

2-1828745 

2-3552617 

2-5472716 

Vi^52l6 

17 

2-2920183 

2-4848011 

2-6927727 

3-15881 

18 

2-4066192 

2-6214652 

2-8543391 

3-^7293 

19 

2-5269502 

2-7656458 

3-0255995 

3-61652 

20 

2-6532977 

2-9177563 

3-2071355 

3-86968 

21 

2-7859625 

3-0782329 

3-3995636 

4-14056 

22 

2-9252607 

3-2475357 

3-6035374 

4-43040 

23 

3-0715237 

3-4261502 

3-8197496 

4-74052 

24 

3-2250999 

3-6145885 

4-0489346 

5-07236 

25 

3-3863549 

3-8133910 

4-29 18707 

5-42743 

26 

3-5556726 

4-0231279 

4-5493829 

5-80735 

27 

3-7334563 

4-2443999 

4-8223459 

6-21380 

28 

3-9201291 

4-4778419 

£-1116867 

6-64883 

29 

4-1661356 

4-7241232 

5-4183879 

7-11425 

30 
31 

4-3219423 

4-9839499 

5-7434912 

7-61225 

4-5380394 

5-258067 1 

6-0881007 

8-14571 

32 

4-7649414 

5-5472608 

6-4533867 

8-71527 

33 

5-0031885 

5-8523600 

6-8405899 

9-32533 

34  , 

5-2533479 

6-1742398 

7-2510253 

9-97811 

35 

5-5160152 

6-5138230 

7-6860868 

10-6765  . 

36 

5-7918101 

6-8720832 

8-147252 

11-4239 

37 

6-0814069 

7-2500478 

8-6360871 

12-2236 

38 

6-3854772 

7-6488004 

9-1542523 

13-0792 

39 

6-7047511 

8-0694844 

9-7035074 

13-9948 

40 
41 

7-0399887 

8-5133060 

10-2857178 

14-9744 

7-3919881 

89815378 

10-9028608 

16-0226 

42 

7-7615875 

9-4755224 

11-5570325 

17-1442 

43 

8-1496669 

9-9966761 

12-2504547 

18-3443 

44 

8-5571502 

10-5464933 

12-9854817 

19-6284 

45 

8-9850077 

11-1265504 

13-7626109 

21-0024 

46 

9-4342581 

11-738^217 

14-5883673 

22-47^ 

47 

9-9059710 

12-3841404 

15-4636693 

£4-0457 

48 

10-4012696 

13-0652681 

10-3914^)94 

25-7289 

49 

10-9213331 

13-7838579 

17-3749788 

27-5wl99  J 

!  50 

11-4673697 

14-5410000 

18-2174775 

2ti-.l570  1 

836 


TABLES. 


TABLE  II. 

SHKW•I^C    THE   PRESENT   VALF«   OF   £l   OR   $ly  DUE   AT   THE   END   Of  AST 
NUMBER   em   YEARS,   F»«M    1    TO   40. 


ys.j  4per  ct.  |4i  per  cfe'  5  perct.  |3i  per  ct.|  6  per  ct.  f  7  perct. 

1 

•961538 

•956938 

•952381 

•947867 

•943396 

•934579 

2 

•924556 

•91573 

•90703 

•898513 

•889996 

•873438 

3 

•888996 

•876297 

•863838 

•851728 

•839619 

•816297 

4 

•454808 

•83856J 

•822702 

•807397 

•792093 

•762895 

6 
~6 

•821927 

802451 

•783526 

•765392 

•747258 

•712986 

•790314 

•767896 

>746215 

•725587 

•70496 

•666042 

7 

•759918 

•734828 

•710681 

•687869 

•665057 

•622749 

8 

•730690 

•703185 

•676839 

652125 

•627412 

•582009 

9 

.702587 

672904: 

.644609 

618253 

•591898 

•543933 

10 
11 

675564 

643928 

•613913 

•586153 

•558394 

•508349 

•649581 

•616199 

•584679 

•573733 

•562787 

•475092 

12 

•624597 

•589664 

•556837 

•526903 

•49696<) 

•444012 

13 

•600574 

•564271 

•530321 

•49958 

•468839 

414964 

14 

•:-77475 

•539973 

•505068 

•473684 

•442301 

.387817 

15 

116 

•555264 

•516720 

481017 

•449141 

417265 

•362446< 

•533908 

•494469 

•458311 

•425979 

393647 

338734 

17 

•513373 

•473176. 

•436297 

•40383 

•371364 

•316574 

18 

•493628 

•4528 

•415521 

•382932 

350343 

•295864 

!l9 

•474642 

•433302 

•395734 

•363123 

•330513 

•276508 

J20 

•456387 

•414643 

•376889 

•344346 

•311804 

•258419 

•438833 

•396787 

•358942 

•326568 

•294155 

241513 

|22 

•421955 

•379701 

•34185 

•309677 

•277505 

•225713 

23 

•405726 

•36335 

•325571 

.293684 

•261797 

•210947 

24 

•390121 

•347703 

•310068 

•278523 

•246978 

•197146 

[26 

26 

375117 

•332731 

•305303 

•26915 

•232998 

•184249 

•360689 

•318402 

•281241 

•250525 

•21981 

•172195 

27 

•340816 

•304691 

•267848 

•237608 

•207368 

•160930 

128 

•333477 

•^91571 

•255094 

•225362 

•19563 

•1*0402 

l29 

•320651 

•279015 

•242946 

•213715 

•184556 

•14056^ 

|30 
131 

•308309 

•267 

•231377 

•202743 

•17411 

•131367 

290460 

•255502 

220359 

•192307 

•164255 

•122773 

j32 

•285058 

•2445 

•209866 

.•182411 

•154957 

•114741 

33 

•274094 

•233971 

•199872 

•173029 

•146186 

•  107234* 

34 

•263552 

•223896 

•190355 

164133 

•137912 

•100219 

35 
36 

254415 

•214251 

•18129 

•155692 
•147399 

•130105 

•093663 

•243669 

•205028 

•172057 

•122741 

•087535 

37 

•234297 

•196299 

•164436 

•140114 

•115795 

•081808 

3tr 

-^25285 

•18775 

•156605 

•132893 

•109182 

•076456 

39 

.216671 

•^179605 

•149148 

•126075 

•103002 

071455 

40i -208289 

•171929 

•142046 

•119608 

•09717 

066780 

TABLES. 

TABLE  III. 


S37 


KEWIIfG    THE   AMOUNT   OF   £  1    OR   $1    ANNUITY    FOR    ANT    NUMBER   OF 
YEARS,    FROM    1    TO    49. 


1 

4  pr.  cent. 

4^  pr.  cent. 

5  pr.  cent. 

5h  pr.  cent, 
1- 

6  pr.  cent. 

7  pr.  cent. 

1- 

1- 

1- 

1- 

1- 

2 

2-04 

2-045 

2-05 

2-055 

£-06 

2-07 

3 

3-1216 

3-137025 

3-1525 

3-16802 

3-1836 

3-2149 

4 

4-246464 

4-278191 

4-31012U 

4-34226 

4-374616 

4-43994 

5 
6 

5-416322 
6-632975 

5-47071 

5-525631 
6-801913 

5-58109 

5-637093 

5-75073 
7-15329 

6-716892 

6-888051 

6-975318 

7 

7-898294 

>  8-0 19 152 

8-14200!i 

8-266894 

8-393837 

8-65402 

8 

9-214266 

9-380014 

9-549109 

9-721573 

9-897467 

10-2598 

9 

10-582795 

10-802114 

11-026564 

1 1-256259 

11-491315 

11-9799 

10 

12.006107 

12-2882 

12-577892 

12-875354 

13-180794 

13-8164 

I] 

13-486351 

13-841179 

14-206787 

14-583498 

14-971642 

15-7836 

12 

15-025805 

15-464032 

15-917126 

16-38559 

16-86994 

17-8884 

13 

16-626838 

17-159913 

17-712983 

18-286798 

18-882132 

30-1406 

\4 

18-291911 

18-932109 

19-59863'i 

20-292572 

21015064 

22-5504 

15 

20-02358{^ 

20-784054 

21-578563 

22-408663 

23-275968 

25-1290 

16 

21-824531 

22-719337 

23-657492 

24-64114 

25-672527 

27-8880 

17 

23-697512 

24-741707 

25-840366 

26-996402 

28-212879 

30-8402 

18 

25-645413 

26-855084 

28-132385 

29-481205 

30-905652 

33-9990 

19 

27-671229 

29-063562 

30-529004 

32-102671 

33-759991 

37-3789 

20 

29-778078 

31-371423 

33-065954 

34-868318 

36-78559 

40-9954 

21 

31-969202 

33-783137 

35-719252 

37-786075 

39-992725 

44-8651 

22 

34-24797 

36-303378 

38-505214 

40-864309 

43-392289 

49-0057 

23 

36-617888 

38-93703 

41-430475 

44-111846 

46-995826 

53-4361 

24 

39-082604 

41-689196 

44-501999 

47-537998 

50-815576 

58-1766 

25 

41-645908 

44-56521 

47-727099 

51-152588 

54-86451 

63-2490 

— 



— 

__. 

— 

26 

44-311745 

47-570645 

51-113454 

54-96598 

59-156381 

68-2490 

27 

47-084214 

50-711324 

54-669126 

58-989109 

63-705763 

74-4838  , 

28 

49-967583 

53-993333 

58-402583 

63-23351 

68-528109 

80-6976 

29 

52-966286 

57-423033 

62-3227 12 

67-711353 

73-639796 

87-3465 

30 

56-084938 

61-007067 

66-438847 

72-435478 

79-058183 

94-4607 

31 

59-328335 

64-752388 

70-76079 

77-419429 

84-801674 

102-073 

32 

62-701469 

68-666245 

75-298829 

82-677498 

90-889775 

110-218 

33 

66-209527 

72*756226 

80-063771 

88-22476 

97-343161 

118-933 

34 

69-857908 

77-030256 

85-066959 

94-077122 

104-183751 

128-258 

35 

73-652225 

81-496618 

90-320307 

100-251363 

111-434776 

138-236 

36 

77-598314 

86-163966 

95-836323 

106-765188 

119-120863 

148-913 

37 

81-702246 

91-041314 

101-628139 

113-637274 

127-268114 

160-337 

38 

85-970.336 

96- 138205 

107-709546 

120-887324 

135-904201 

172-561 

130 

.90-40915 

101-464424 

114-095025 

128-5361 27 

145-053453 

185-640 

Uo 

95-025516 

!  07-030323 

120-799774 

136-6056141 1.^4-761961 

199-635 

T  t 


TABLLS. 


[lEWING  Tilt  MIT. -LA 


TABLE  IV. 

AVollllI  OF  £\  OR  ${  AXNUIIY,  TOR  AJfY  KtTMBfiE 
OF  YFARS  FROM  1  TO  40. 


ys,|4  per  ceut.l  4^j  per  ct,  |.>  per  ceiit..(  5J.  per  ct.  |6  per  cent.|  7  per  ct.  j 

1 

0,il(il54 

0,95694 

0,95238 

0,94786 

0,94339|  0,9346! 

o 

1,88609 

1,87267 

1,85941 

1  8463 

1,83339! 

1,8080 

3 

2,77509 

2,74896 

2,72326 

2,6979 

2,67301 

2,6243 

4 

3,62989 

3,58752 

3,64696 

3,49862 

3,4651 

3,3872 

i~6 

4,45182 
6,24214 

4,38997 

~5^57"87 

4,32948 
"'5767"669 

4.26759 

4,21236 
4,92732 

4,1001 
4,7666 

4,97699 

1  7 

6,00205 

6,8927 

5,78637 

5,66888 

6,68238 

6,3892 

8 

6,73274 

6,59589 

6,46321 

6,30522 

6,20979 

6,9712 

p 

7,43533 

7,26879 

7,10782 

6,91786 

6,80169 

6,5152 

10 

n 

8,11089 

7,91272 
8,52892 

7,72173 

7,49866 

7,36008 

7,0235 

8,76048 

8,30641 

8,04898 

7,88687 

7,4986 

!l2 

9.38607 

9,11858 

8,86326 

8,6707 

8,38384 

7,9426 

13 

9,98566 

9,68285 

9,39367 

9,06522 

8,86268 

8,3576 

14 

10,56312 

10,22282 

9,89864 

9,63396 

9,29498 

8,7454 

jl5 

m 

11,11839 

10,73954 

10,37966 

9,97824 

9,71225 

9,1079 

11,65229 

11,23401 

10,83777 

10,39936 

10,10589 

9,4466 

17 

12,16567 

11,70719 

11,27407 

10,79852 

10,47726 

9,7632 

18 

12,65929 

12,15099 

11,68958 

11,17687 

10,8276 

10,059 

19 

13,13394 

12,69329 

12,08632 

11,63549 

11,16811 

10,335 

20 
¥l 

13,59032 

13,00793 

12,46221 

11,87641 

11,46992 
11,76407 

10,694 

14,02916 

13,40472  12,82115 

12,1976 

10,835 

22 

14,45111 

13,79442 

13,163 

12,50299 

12,04158 

11,061 

23 

14,85684 

14,14777 

13,48807 

12,79246 

12,30338 

11,272 

24 

15,24696 

14,49648 

13,79864 

13.06682 

12,65035 

11,469 

25 

15,62208 

14.82821 

14,09394 

13.3688 

12,78336 

11,653 

26 

15,98277 

15,14661 

14,37518 

13,67338 

13,00316 

11,826 

27 

16,32959 

1^,4513 

14,64303 

13,80702 

13,21053 

11,986 

28 

16,66306 

15,74287114,89813 

14,02848 

13,40616 

12,137 

29 

16,98371 

16,02189 

15,14107114,23838 

13,69072 

12,277 

30 
37 

1 7,20202 

16,28889 

15.37246 
T5^92¥l 

14,43733 
T4T6259~ 

13,76483 
13,92908 

12,409 

17,58349 

12,631 

32 

17,87355 

16,78889 

15,80268 

14,80463 

14,08398 

12,646 

33 

18,14764 

17,02286 

16,00256 

14,97404 

14,22917 

12,753 

34 

18,4112 

17,24676 

16,1929 

15,13461 

14,36613 

12,864 

35 
,36 

18,66461 
18»,9082T; 

17,46101 

17,66604 

16,37419 

15,2868 

14,49533 

12,947 

16,54086 

15,43106 

14,61722 

13,035 

37 

19J4258 

17,86224 

16,71129 

16.66779 

14,73211 

13,117 

38 

19,36787 

18,04999 

1G,86789|  16,6974 

14,84048 

13,193 

,30 

19,58448 

18,22965 

17,01^704 

16,82024 

14,9427 

13,264 

|40 

19.79277 

18,40158 

17.16909 

15.93667 

15.03913 

13,331 

TABLES. 


339 


TABLE  V. 

THE   ANJfUITY   WHICH   £l   OR    $1  WILL   PURCHASE   FOR   ANY   NUMBER   OF 
YEARS    TO    C03IE,   FROM    1    TO    40. 


1 

4  per  ct. 
1,04 

4^  per  ct. 
1,045 

5  per  ct. 
1,05 

54  per  ct. 
1,055 

6  per  ct. 
1,06 

7  per  ct^ 

1,07 

2 

,5302 

,534 

,5378 

,54162 

,54544 

,55309 

3 

,36035 

,36377 

,36721 

,37065 

-  ,37411 

,38105 

4 

,27549 

,27874 

,28201 

,28582 

,28859 

,29523 

5 

6 

,22463 
,19076 

,22779 

,23097 
,19702 

,23487 
,20092 

,23739 

,24389 

,19388 

,20336 

,20977 

7 

,16661 

,1697 

,17282 

,17671 

,17913 

,18556 

8 

,14853 

,15161 

,15473 

,15859 

,16103 

,16747 

9 

,13449 

,13757 

,14069 

,14455 

,14702 

,15349 

10 

n! 

,12329 
,11415 

,12638 

,1295 
,12039 

,13334 
,12424 

,13587 
,12679 

,14238 

,11725 

,13336 

12 

,10655 

,10967 

,11282 

,11667 

,11927 

,12592 

13 

,10014 

,10327 

,10645 

.11031 

,11296 

,11965 

14 

,09467 

,09782 

,10102 

,10489 

,10758 

,11435 

15 

,08994 

,09311 

,09624 

,10022 

,10296 

,10979 

16 

,08582 

,08901 

,09227 

,0962 

,09895 

,10586 

17 

,0822 

,08542 

,0887 

,0926 

,09544 

,10243 

18 

,07899 

,08224 

,08555 

,08947 

,09235 

,09941 

19 

,07614 

,07941 

,08274 

,08699 

,08962 

,09676 

20 
21 

,07359 
,07128 

,07688 

,08024 

,08427 

,08718 

,09439 

,0746 

,078 

,08198 

,085 

,09230 

22 

,0692 

,07254 

,07597 

,07998 

,08303 

,09041; 

23 

,06731 

,07068 

,07414 

,07825 

,08128 

,08880* 

24 

,06559 

,06899 

,07247 

,07653 

,07968 

,08719' 

25 

2G 

,06401 

,06744 

,07095 

,07503 
,07367 

,07823 
,0769 

,0858  Ij 

,06257 

,06602 

,06956 

,08457} 

27 

,06124 

,06472 

,06829 

,07242 

,0757 

,083431 

28 

,06001 

,06352 

,06712 

,07128 

,07459 

,082391 

29 

,05888 

,06241 

,06604 

,07023 

,07358 

,08145; 

30 
31 

,05783 

,06139 

,06506 

,06926 

,07272 

,08058 

,05685 

,06044 

,06413 

,06837 

,07179 

,07980 

32 

,05595 

,05956 

,06328 

,06754 

,071 

,07908 
,07841 

33 

,0551 

,05874 

,06249 

,06678 

,07027 

34 

,05431 

,05798 

,06175 

,06607 

,06959 

,07779 

35 

,05358 

,05727 

,06107 

,06541 

,06899 

,07724 

36 

,05289 

,0566 

,06043 

,0648 

,06839 

,07672 

37 

,05224 

,05598 

,05984 

,06423 

,06785 

,076241 

38 

,05163 

,0554 

,05928 

,0637 

,06735 

,07579 

39 

,05106 

,05485 

,05876 

,06321 

,06689 

,07539 
,07501 

40 

,05052 

,05434 

,05828 

,06274 

,06646 

o4tf 


TABLEfe-. 


TABLE  VL 

VALUE  OF  AN  ANNUITY  OF  j£l  OR  $1,  AT  DIFFERESTT  RATES  PER  CENT. 
PAYABLE  YEARLY,  HALF  YEARLY,  aUARTERLY,  DAILY  OR  MOMENTLY, 
rOR  EVER. 


Rate 
per 
cent, 

3 

H 

4 

4i 

5" 

6 

^^ 

7 

Perpetuity 
payable 
yearly. 

Half 

yearly. 

Quar- 
terly. 

Daily. 

Perpetuity 

payable 

momently. 

33,3333 
28,5714 
25,0000 
22  2*^22 
2o!oOOO 
18,1818 
16,6666 
15,5556 
14,2857 

33,6022 
28,7886 
25,2525 
22,4699 
20,2429 
18,4298 
16,9147 
15,6299 
14,5349 

33,6927 
28,9083 
25,3807 
22,5938 
20,3583 
18,5529 
17,0380 
15,7550 
14.6575 

33,8238 
29,0841 
25,4858 
22,7175 
20,4763 
18,6630 
17,0554 
15,7743 
14,7694 

33,8308 
29,2908 
25,4992 
22,7244 
20,4959 
18,6812 
17,1635 
15,7902 
14,7800 

J\'ote.     This  Table  is  calculated  by  the  Rule,  Case  I.  of 
basing  Annuities  for  ever." 


Pur^ 


TABLES. 


341 


TABLE  VII. 

VALUE    OE   AN   ANNUITY   OF   £l    OR   $1,   FOR   A  SINGLE   LIFE. 


3  per 

3^  per 

4  per 

4i  per 

5  j)er 

6  per 

Age. 

cent. 

cent. 

cent. 

cent. 

cent. 

cent. 

10 

19,87 

18,27 

16,88 

15,67 

14,60 

12,80 

12 

19,60 

18,05 

16,69 

15,61 

14,47 

12,70 

16 

19,19 

17,71 

16,41 

15,27 

14,27 

12,55 

18 

18,76 

17,33 

16,10 

15,01 

14,05 

12,48 

20 

18,46 

17,09 

15,89 

14,83 

13,89 

12,30 

22 

18,15 

16,83 

15,67 

14,64 

13,72 

12,15 

25 

17,66 

16,42 

15,31 

14,34 

13,46 

12,00 

28 

17,16 

15,98 

14,94 

14,02 

13,18 

11,75 

30 

16,80 

15,68 

14,68 

13,79 

12,99 

11,60 

33 

16,25 

15,21 

14,27 

13,43 

12,67 

11,35 

35 

15.86 

14,89 

13.98 

13,17 

12,45 

11.15 

38 

15,29 

14,34 

13,52 

12,77 

12,09 

10,90 

40 

14,84 

13,98 

13,20 

12,48 

11,83 

10,70 

43 

14,19 

13,40 

12,68 

12.02 

11,43 

10,35 

45 

13,7S 

12,99 

12,30 

11,70 

11,14 

10,10 

48 

13,01 

12,36 

11,74 

11,19 

10,68 

9,75 

50 

12,51 

11,92 

11,34 

10,82 

10,35 

9,45 

53 

11,73 

11,20 

10,70 

10,24 

9,82 

9,00 

55 

11,18 

10,69 

10,24 

9,82 

9,44 

8,70 

50 

10,32 

9,91 

9,52 

9,16 

8,83 

8,20 

60 

9,73 

9,36 

9,01 

9,69 

8,39 

7,80 

63 

8,79 

8,49 

8,20 

7,94 

7,68 

7,20 

65 

8,13 

7,88 

7,63 

7,39 

7,18 

6,75 

6P 

7,10 

6,91 

6,75 

6,54 

6,36 

6,00 

.70 

6,38 

6,22 

6,06 

5,92 

5,77 

5,50 

73 

5,25 

5,14 

5,02 

4,92 

4,82 

4,60 

75 

4,45 

4,38 

4,29 

4,22 

4,14 

4,00 

77 

3,63 

3,57 

3,52 

3,47 

3,41 

3,30 

79 

2,78 

2,74 

2,70 

2,67 

2,64 

2,25 

80 

2,34 

2,31 

2,28 

2,26 

2,23 

2,15 

This  Table  is  formed  by  ascertaining  from  Bills  of  mortality  ths 
mean  Itngth  of  the  lives  of  persons  of  a  certain  age,  and  then  cal- 
culating the  value  of  the  annuity  for  the  number  of  years  they  may 
thus  be  expected  to  live.  This  mean  is  called  the  Expectation  ot" 
Life,  at  any  given  age,  which  is  exhibited  in  the  following  table. 


:42 


CIRCULATING  DECIMALS. 
TABLE  VIIL 


^EXPECTATION    OF    LIFE    AT 

SEVERAL   AGES. 

Age, 

1 

Expec. 

Age. 
20 

Expec. 

Age. 

42 

Expec. 

Age. 
62 

Expec. 

Age. 
82 

Expec. 

29,80 

33,62 

21,65 

11,68 

3,31 

2 

37,92 

22 

32,46 

45 

20,10 

65 

10,20 

85 

2,64 

3 

40,18 

25 

30,83 

47 

19,07 

67 

9,20 

87 

2,40 

4 

41,32 

27 

29,74 

50 

17,55 

70 

7,80 

90 

2,12 

6 

41,77 

30 

28,12 

52  16,55 

72 

7,00 

92 

1,60 

8 

41,34 

32 

27,04 

55 

15,10 

75 

5,84 

93 

1,14 

10 

40,25 

35 

25,34 

57 

14,12 

77 

6,15 

94 

0,90 

12 

38,57 

37 

24,30 

59 

13,15 

79 

4,40 

96 

0,66 

16 

36,64 

40 

22,82 

60 

12,66 

80 

4,00 

96 

0,50 

18 

34,77 

i 

CIRCULATING  DECIMALS 

ARE  produced  from  Vulgar  Fractions,  whose  denominators  do 
not  measure  their  numerators,  and  are  distinguished  by  the  con- 
tinual repetition  of  the  same  figures. 

1.  The  circulating  figures  are  called  rrpetends ;  and,  if  one  figure 
only  repeats,  it  is  called  a  sn?«-/e  rcpetend :  As    1111,  &c.  '6666,  &c. 

2.  A  compound  repeicnd  has  the  same  figures  circulating  alter- 
nately :  As  -OlOtOi,  &c.  -379379379,  &c. 

3.  if  other  figures  arise  before  those  which  circulate,  the  deci- 
mal is  called  a  mixed  repetend  ;  thus,  '375565,  &c.  is  a  mixed  single 
repeiend^  and  '378123123,  he.  a  mixed  compound  repetend. 

4.  A  single  repetend  is  expressed  by  writing  only  the  circula 
ting  figure  with  a  point  over  it;  thus,  -1111,  &c.  is  denoted  by 

•1,  and  -6666,  &c.  by  -6. 

5.  Compound  repetends  are  distinguished  by  putting  a  point  over 
the  first  and  last  repeating  figures  ;  thus,  -010101,  &c.  is  writtea 

•01,  and  -379379379,  &c.  thus,  -379. 

6.  Similar  circulating  decimals  are  such  as  consist  of  the  same 
number  of  figures,  and  begin  at  the  same  place,  either  beture  ur 

after  the  decimal  point ;  thus,  '3  and  '5  are  similar  circulates  ;  as 

are  also  3-54  and  7*36,  &:c. 

7.  Dissimilar  repetends  consist  of  an  unequal  number  of  figures, 
and  begin  at  different  places. 

C.  Similar  and  conterminous  circulates  are  such  as  begin  and  end 


at  the  same  place  ;  as  4734576,  9w3528  and  -05463,  kc. 


REDUCTION  OF  CIRCULATING  DECIMALS.       345 
REDUCTION  OF  CIRCULATING  DECIMALS. 

CASE  I. 

To  reduce  a  simple  Repeiend  to  its  equivalent  Vidgar  Fraction. 

Rule.* 

1.  Make  the  given  decimal  the  numerator,  and  let  the  denomi- 
nator be  a  number,  consisting  of  so  many  nines  as  there  are  re- 
curring^ places  in  the  repetend. 

2.  If  there  be  integral  figures  in  the  circulate,  so  many  cyphers 
must  be  annexed  to  the  numerator  as  the  highest  place  of  the 
repetend  is  distant  from  the  decimal  point. 

Examples. 

1.  Required  the  least  vulgar  fractions  equal  to  -3  and  -324. 

•    •3=1=1 ;  and  .324=A|4.==.r..     Ans.  -i-  and  if. 
?,  Reduce  -7  to  its  equivalent  vulgar  fraction.  Ans.  ^. 

3.  Reduce  2.37  to  its  equivalent  vulgar  fraction.      Ans.  VsV* 

4.  Required  the  least  vulgar  fraction  equal  to  -384615.  Ans.  y^. 

CASE  H. 

To  reduce  a  mixed  Repetend  to  its  equivalent  Vulgar  Fraction. 

RULE.J 

1.  To  so  many  nines  as  there  are  figures  in  the  repetend,  annex 
«o  many  cyphers  as  there  are  finite  places,  (that  is,  as  there  are 
decimal  places  before  the  repeiend)  for  a  denominator. 

"'■  If  unily,  with  cyphers  annexed,  be  divided  by  9  ad  infinitum^  the  quotient 
v'ill  be  1  continually ;  that  is,  if  i  be  reduced  to  a  decimal,  it  will  produce  the 

circulate  •!,  and  since  •  1  is  the  decimal  equivalent  to  ^,  -2  vnW  =?.,  3=3,  and  so 

on  till  'O^-g^l .  Therefore  every  single  repetend  is  equal  to  a  vulgar  fraction, 
whose  numerator  is  llie  repeating  figure  and  denominator  9. 

Again,  J.-  or  -1-  being  reduced  to  decimals,  make  -010101, &c.  and  -001001001, 

k    ^'c.  ad  inJmitumr=z-0\  and  -001 ;  tliat  is,  9V="01i and  ^ly  =-001,  consequently 

/,,-~-02,  r;\— -03,  kc.  and  g  fg  =-002,  g fg-  =:-003,  &c.  and  the  same  will  hold 
universally. 

t  In  like  manner  for  a  mixed  circulate ;  consider  it  as  divisible  i»to  its  finite 
and  circulating  parts,  and  the  same  principle  will  be  seen  to  run  through  them 

also;  thus  the  mixed  circulate  -13  is  divisible  into  the  finite  decimal  -1,  and  the 

repetend  -03  :  but  '1— jVi  ^^<^  *03  would  be  equal  to  f  provided  the  circulation 
began  immediately  after  the  place  of  units  ;  but  as  it  begins  after  the  place  of 
tenths,  it  is  |  of  _.i_— JL,  and  so  the  vulg;ar  fraction='13  is  _.>_-|-_3_=r  .»„-|-_a. 
=-j}|-,  and  id  the  same  as  by  the  rul'?. 


344       REDUCTION  OF  CIRCULATING  DECIMALS. 

2.  Multiply  the  nines  in  the  saifJ  denominator  by  the  finite  pail, 
and  add  the  repeating  decimals  to  the  product  for  the  numerator. 

3.  If  the  repetend  begins  in  some  integral  place,  the  finite  value 
of  the  circulating  part  must  be  added  to  the  finite  part. 

Examples. 

1.  What  is  the  vulgar  fraction  equivalent  to  '153  ? 

There  being  1  figure  in  the  repetend,  and  2  finite  places,  I  an- 
nex 2  cyphers  to  9  for  a  denominator,  viz.  900;  then  1  multiply 
the  9  in  the  denominator  by  the  two  figures  in  the  finite  part,  and 
add  the  repeating  figure  for  a  numerator;  thus,  9xl5-f 3=138 
numerator. 

Therefore,  'IGS— if^==y2_;L  the  Ans. 

2.  What  is  the  least  vulgar  fraction  equal  to  -4123  ?     Ans.  ||^|. 

3.  Required  the  finite  number  equivalent  to  45-78  ?  Ans.  45||,i 

CASE  HI. 

To'  make  any  number  of  dissimilar  repctends  similar  and  coniermi 
nous ;  that  is,  of  an  equal  number  of  places. 

Rule.* 
Change  them  into  other  repetend.e,  which  shall  each  consist  of 
so  many  figures,  as  the  least  common  multiple  of  the  sums  of  the 
several  numbers  of  places,  found  in  allthe  repetends,  contains 
units. 

Examples. 

1.  Make  6  317  ;  3-45  ;  62-3  ;  191-03  ;  -057  ;  5-3  and  1-359  sim- 
ilar and  conterminous. 

Here,  in  the  first  repetend,  there  are  three  places,  in  the  se- 
cond, one,  in  the  third,  none,  in  the  fourth,  two,  in  the  fifth,  three, 
iM  the  sixth,  one,  and  in  the  seventh,  one. 

Now  find  the  least  common  multiple  of  these  several  sum«,  thus  : 
3\.3,  1,  2,  3,  1,  1 
j  „  ^^^  2x3=6  units  ;    thereforp,  the  similar  and 

conterminous  repetends  must  contain  6  place.^.j 

=*  All)''  gjivcn  repetend  whatever,  whether  sing^le,  compoun'l,  pure,  or  inixed, 
itiuy  be  triinsformcd  into  another  repefend,  which  shall  coir  ii^t  of  an  equal 

ur  gi-enlri-^iiiiiil-ci-  of  figures  at  plcatiure  ;  thus,  •.>  may  be  transformed  into  '33, 

or  -333,  ^c.  also  -TOr-'TinO-r-v-TOT,  and  so  on. 

t  The  learner  may  observe  that  the  similar  and  contermiuous  repetends  be- 
^j-in  just  so  far  from  unity,  as  is  the  farthest  among;  the  dissimilar  repetends ;  and 
il.  is  £0  in  all  cases. 


REDUCTION  OF  CIRCULATING  DECIMALS,       345 

Dissimilar  made  similar  and  conterminous, 

6-317=     631731731 

3<45  =  3  46655555 

52-3  =  62  30000000 

191-03  —191  03030303 

•057=        •05705705  ' 

5-3     =     5-33333333 
1-369=     1-3599999:9 
}.       2.  Make  '531,  -7348,  -07  and   0603  similar  and  conterminous. 

CASE  IV. 

To  find  whether  the  decimal  fraction,  equal  to  a  given  vulgar  one, 
be  finite  or  infinity,  and  how  many  places  the  repeiend  will  con- 
sist of. 

Rule* 

1.  Reduce  the  given  fraction  to  its  least  terms,  and  divide  the 
denominator  by  2,  5  or  10,  as  often  as  possible. 

2.  Divide  9999,  &;c.  by  the  former  result,  till  nothing  remain, 
and  the. number  of  9s  used  will  show  the  number  of  places  in  the 
repetend  ;  whicii  will  begin  after  so  many  places  of  figures  as 
there  were  10s,  2s,  or  5s,  divided  by. 

If  the  whole  denominator  vanish  in  dividing  by  2,  5  or  10,  the 
decimal  will  be  finite,  and  will  consist  of  so  many  places  as  you 
perform  divisions. 

*  In  dividing  1-000,  &c.  by  any  prime  number  whatever,  except  2  or  5,  the 
figures  in  the  quotient  will  begin  to  repeat  over  again  as  soon  as  the  remainder 
is  1 :  and  since  999,  &:c.  is  less  than  1000,  &:c.  by  1,  therefore  999,  &c.  divided 
by  any  number  whatever,  will,  when  the  repeating  figures  are  at  their  period, 
leave  0  for  a  remainder. 

Now,  whatever  number  of  repeating  figures  we  have,  wdien  the  dividend  is  1, 
there  will  be  exactly  the  same  number,  when  the  dividend  is  any  other  number 
whatever. 

Thus,  let  -390539053905,  &c.  be  a  circulate,  whose  repeating  part  is  3905. 
Now,  every  repetend  (3905,)  being  equally  multiplied,  must  give  the  same  pro- 
duct :  For  although  these  products  will  consist  of  more  places,  yet  the  overplus 
in  each,  being  alike,  will  be  carried  to  the  next,  by  which  means,  ea^  product 
will  be  equally  increased,  and  consequently  every  four  places  will  continue  alike. 
And  the  same  will  hold  for  any  other  number. 

Now  from  hence  it  appears  that  the  dividend  may  be  altered  at  pleasure,  and 

the  number  of  places  in'  the  repetend  will  still  be  the  same ;  thus,  -J_".-09  ;  and 

fi  ^^  -]\-X4;--3C,  whence  ihc  number  of  pl.irc:,  in  each  arc  rililtc 

r  .1 


346         ADDITION  OF  CIRCULATING  DECIMALS. 

Examples. 
1.  Required  to  find  whether  the  decimal  equal  to  2W0  ^^^  finite 
or  infinite,  and  if"  infinite,  how  many  places  that  repetend  will  con 
sist  of. 

475  19       \  (2)    (2)    (2) 

:56=28=14=7. 


\  475  19      \  ( 

First  25  ) ^ 2  ]\n=l 

/2800        112     / 


7)999999 
Then,      Taq^^t '^  therefore,  because  the  denominator  112  did 

not  vanish  in  dividing^  by  2,  the  decimal  is  infinite,  and,  as  six  9s 
were  used,  the  circulate  consists  of  6  places,  beginning  at  the  fifth 
place,  because  four  2s  were  used  in  dividing. 

2.  Let  ji  be  the  fraction  proposed. 

3.  Let  f  be  the  fraction  proposed. 

ADDITION  OF  CIRCULATING  DECIMALS. 

Rule. 

1.  Make  the  repetends  similar  and  conterminous,  and  find  their 
sum  as  in  common  addition. 

2.  Divide  this  sum  (of  the  repetends  only)  by  so  many  nines  as 
there  are  places  in  the  repetend,  and  the  remainder  is  the  repe- 
tend of  their  sum  ;  which  must  be  set  under  the  figures  added, 
with  cyphers  on  the  left  hand,  when  it  has  not  so  many  places  as 
the  repetends. 

3.  Carry  the  quotient  of  this  division  to  the  next  column,  and 
proceed  with  the  rest  as  infinite  decimals. 

Examples. 

1.  Let  5'3+59'435C+397  6+519+-39+217'5  be  added  together, 

5  3       =     6-3333333 

59'4366=  59-435G356 

397-6       =:397-360GGG6 

519-         —519-0000000 

•39     ~       -3939393 

^n7-5      .=217-6555555 


1199-3851303 


1199  3851305  the  sum. 
in  this  question,  the  sum  of  the  repetends  is  2851303,  which 
divided  by  999999,  gives  2  to  carry  to  the  next  column  5,3,0,  &c. 
and  the  remainder  is  851305. 


xMULTIPLlCATION  OF  CIRCULATING  DECIMALS.   3  n 

2.    Let   3275-3194-36-454-123-19+5-3173-f  n2-3513+lM31 
+  •125+29-10053  be  added  together.  Ans.  3593-00042. 

SUBTRACTION  OF  CIRCULATING  DECIMALS. 

Rule. 
Make  the  repetends  similar  and  conterminous,  and   subtract  as 
Tisual,  observing,  that  if  the  repetend  of  the  number  to  be  sub- 
iracted  be  greater  than  (he  repetend  of  the  number  it  is  to  be  ta- 
ken from,  then  the  right  hand  of  the  remainder   must  be   loss  by 
inity  than  it  would  be  if  tlie  expressions  were  finite. 

Examples. 

1.  From  5703  take  29-73587 

5703       =5703030 

23-73587=29-73587 


27.29442  ^he  difference. 
2.  From  325  17  take  137-5819.  Ans.  1 87-5957. 

MULTIPLICATION  OF  CIRCULATING  DECIMALS, 

Rule. 

1.  Turn  both  the  terms  into  their  equivalent  vulgar  fractions, 
and  find  the  product  of  those  fractions  as  usual. 

2.  Turn  the  vulgar  fraction  expressing  the  product,  into  an 
equivalent  decimal  one,  and  it  will  be  the  product  required. 

Examples. 

1.  Multiply  -54  l)v  -15..     'b\=l^^j\  and  •15=ii=J7_ 
■ixXi'E—ih—'^^^  ^he  product. 

2.  Multiply  378-5  by  236.  Ans.  8959-148. 

DIVISION  OF  CIRCULATING  DECIMALS. 

y 

Rule. 

1.' Change  both  the  divisor  and  dividend  into  their  equivalent 
vulgar  fractions,  and  find  their  quotient  as  usual. 

2.  Turn  the  vulgar  fraction  expressing  the  quotient,  into  its 
equivalent  decimal,  and  it  will  be  the  quotient  required. 


348 


ALLIGATION. 

Examples. 


1.  Divide  '54  by  '15. 


•B4=5A=:=:_6^  an^  •13=11=  "L 


it-^4V=AXV=Vt  =3ff=:3'606493  the  quotient. 
2.  Divide  3458  by  6.  Ans.  618  S3. 


ALLIGATION 


IS  the  method  of  mixing  two  or  more  simples  of  different  quali- 
ties, so  that  the  composition  may  be  of  a  mean  or  middle  quahty  ; 
It  consists  of  two  kinds,  viz.  Alligation  Medial,  and  Alligation  Al- 
ternate. 

ALLIGATION  MEDIAL 

Is,  when  the  quantities  and  prices  of  several  things  are  given, 
to  find  the  mean  price  of  the  mixture  compounded  of  those  things. 

Rule. 
As  the  sum   of  the   quantities,  or  the  whole  composition,  ^s  to 
their  total  value  ;  so  is  any  part  of  the   composition  to   its  mean 
price  or  value. 

Examples. 
1.  A  Tobacconist  would  mix  60lfe  of  tobacco,  at  6d.  per  ft  with 
50ife  at  Is.  40ife  at  Is.  6d.  and  30ife  at  2s.  per  ft:    What  is   1ft  of 
this  mixture  worth  ? 

ft         s.    d.      £   s.  ft       £     ft         ■ 

60  at  0     6  is   1    10     As  180  :  10  :  1 
60  -_  1     0  —  2   10  1 

40  —  1     6  —  3     0  — 

30—2     0  —  30 


Sum   of  the  ?  io^t  «  i      i     i.. 
simples,      M80  Total  value  10 


0 


10 
20 

180)200(13. 
180 


20 
12 


180)240(lid.  s.  d. 


180  Ans.  1 


CO 


ipr.ft 


2.  A  farmer  would  mix  20  bushels  of  wheat  at  gl  per  bushel, 
IG  bushels  of  rye  at  75c.  per  bushel,  12  bushels  of  barley  at  50c. 


ALLIGATION  ALTERNATE.  349 

per  bushel,  and  8  bushels  of  oats  at  40c.  per  bushel ;  What  is  the 
value  of  one  bushel  of  this  mixture  ?  Ans.  73c.  5|m. 

3.  A  wine  merchant  mixes  12  gallons  of  wine,  at  75c.  per  gal 
Ion,  with  24  gallons  at  90c.  and    IG  gallons  at  §1   10c.  :   What  is  a 
gallon  of  this  composition  worth  ?  An?.  92c.  6m. 

4.  A  goldsmith  melted  together  8oz.  of  gold  of  22  carats  fine, 
1^  life  8oz.  of  21  carats  fine,  and  lOoz.  of  18  carats  fine  :  Pray  what 
I  is  the  quality,  or  fineness  of  the  composition  ? 

^^2+Yi^J^lj  carats  fine,  Ans. 

8+20+10 

5.  A  refiner  meits  51b  of  gold  of  20  carats  fine  ^ith  8ft  of  1^ 
carats  fine  :  How  much  alloy  must  be  put  to  it,  to  make  it  22  ca- 
rats fine  ?  r=r 

.  22— 6X20+8X1  S^i-S+S^Sy^^. 
Answer.     It  is  not  fine  enough  by  o^\  carats,  so  that  no  alloy 
i^   must  be  added,  but  more  gold. 

ALLIGATIOjY  ALTERjYATE'^ 

Is  the  method  of  finding  ivhat  quantity  of  each  of  (he  ingredi- 
ents, whose  rales  are  given,  will  compose  a  mixture  of  a  given 
rate  :  So  that  it  is  the  reveise  of  Alligation  Medial,  and  may  he 
proved  by  it. 

CASE  I. 

The  whole  work  of  this  case  consists  in  linking  the  extremes 
truly  together  and  taking  the  differences  between  them  and  tho 
mean  price,  which  differences  are  the  quantities  sought. 

Rule. 
1.  Place  the  several    prices   of  the   simples,  being  reduced  t© 
one  denomination,  in  a  column  under  each  other,  the  least  upper- 
most, and  so  gradually  downward,  as  they  increase  with  a  line  oi' 

■  Demon.  By  connecting  the  less  rate  willi  the  greater,  and  placing  the  dif- 
Jcreace  between  them  and  the  mean  rate  alternately,  or  one  after  the  other  ui 
turn,  the  quantities  resulting  nre  such,  that  there  is  precisely  as  much  gained  by 
one  quantity  as  is  lost  by  the  other,  and  therefore  the  gain  and  loss,  upon  the 
whole,  are  equal,  and  are  exactly  the  proposed  rate.- 

In  like  manner,  let  the  number  of  simples  be  what  it  may,  and  with  how  many 
soever,  each  one  is  linked,  since  it  is  always  a  less  with  a  greater  than  the  mcau 
price,  there  will  be  an  equal  balance  of  loss  and  giiin  between  every  two,  anc' 
consequently  an  equal  balance  on  the  whole. 

It  is  obvious  from  the  rule,  that  questions  of  this  sort  admit  of  a  great  variety 
of  answers ;  for  having  found  one  answer,  we  may  find  as  many  more  as  we 
please,  by  only  multiplying  or  dividing  each  of  tlie  quantities  found,  by  2,  3,  4, 
&c.  the  reason  of  which  is  evident ;  for  if  two  quantities  of  two  simples  make 
a  balance  of  loss  and  gain  with  respect  to  the  mean  price,  so  must  also  the  dou- 
ble or  triple,  the  half  or  third  part,  or  any  other  ratio  of  these  quantities,  and 
so  on  ad  infinitum. 

If  any  one  of  the  simples  be  of  little  or  no  value  with  respect  to  the  rest,  ifs 
rate  is  supposed  to  be  nothing,  as  water  mix^d  with  wine,  and  alloy  with  gold 
}ind  silver. 


obO 


ALLIGATION  ALTERNATE. 


connection  at  the  left  hand,  and  the  mean  price  at  the  icl't  hand 
of  all. 

2.  Connect,  with  a  continued  line,  the  price  of  each  simple,  or 
ingredient,  which  is  less  than  that  of  the  compound,  with  one  or 
any  number  of  those  which  are  greater  than  the  compound,  and 
each  greater  rate  or  price  with  one  or  an}'  number  of  the  less. 

3.  Place  the  difference,  between  the  mean  price  (or  mixture 
rate)  and  that  of  each  of  the  simples,  opposite  to  the  rates  with 
which  they  ai^  connected. 

4.  Then,  if  only  one  difference  stand  against  any  rate,  it  will 
be  the  quantity  belonging  to  that  rate  ;  but  if  there  be  several, 
their  sum  will  be  the  quantity. 

.  Examples. 
1.  A  merchant   has  spices,  some  at  Is.  6d.  per  Jfe,  some  at  2s. 
some  at  4s.  and  some  at  5s.  per  fe  :  How  much  of  each  sort  must 
he  mix  that  he  may  sell  the  mixture  at  3s.  4d.  per  3fe '{ 


Mean 
rate40d 


28 
38 
16 
04-20 
.20 
22 

22-fl6     oo 
28  at  J  6 
28  —  2  0 
38  —  4  0, 
J8  — 5  0 

Note.  I'lie-e  seven  answers  arise  from  as  many  various  ways 
of  linking  the  ralt'«  of  the  ingredients  together. 

2.  *A  fnerchant  has  Canary  wine,  at  3s.  per  gallon,  Sherry,  at 
23.  Id.  and  Claret  at  Is.  5d.  per  gallon  :  How  much  of  each  sort 
must  he  take,  to  sell  it  at  2s.  4d.  per  gallon  ? 


Meanratc28d. 


^36'*.   3f  n   14  at  3  0) 

]d.  {  25^)  8  8       2    1) 

f  17-^  8  8        15) 


per  gallon. 


*  Note,  the  2d  and  3(1  qnctiom  admit  but  of  one  way  of  linking;,  and  po 
but  of  one  answer  ;  yet  all  numbers  in  the  same  proportion  between  tliemselvc.-:, 
as  the  numbers  which  compose  the  answer,  will  likewise  satisfy  the  condition  of 
the  question. 


ALLIGATION  ALTERNATE. 


o.  How  ranch  barley  at  40c.  rye  at  60c.  and  wheat  at  80c.  per 
bushel,  must  be  mixed  together,  that  the  compound  may  be  worth 
62iG.  per  bushel? 

Ans.  17|  busfiels  of  barley,  Ilk  ofr^e,  and  25  of  wheat. 

4.  A  gohlsmilh  woukl  mis  gohl  o(^l9  carats  fine,  witli  some  of 
16,  18,  23  and  24  carats  tine,  so  that  the  compound  may  be  21  ca- 
rats fine  :    What  quantity  of  each  must  he  take  ? 

Ans.  5oz.  of  16  carats  tine,  3oz.  of  18,  3oz.  of  t^,  lOoz.  of  23, 
and  lOoz.  of  24  carats  line. 

5,  It  is  required  to  mix  several  sorts  of  wine,  at  60c.  90c.  and 
^1  I5c.  per  gallon,  with  water,  that  the  mixture  may  be  worth  Idc. 
per  gallon  :  Of  how  much  of  each  sort  must  the  composition  con- 
sist ? 

Ans.  40gall8.  of  water,  15galls.  of  wine,  at  60c.  ISgalls.  do.  at 
90c.  and  75galls.  do.  at  gl  15c. 


(?ASE 


II. 


IVhen  the  rates  of  all  the  ingredients,  the  quantity  of  hut  one  of  them, 
and  the  mean  rate  of  the  zvhole  mixture  arc  given,  to  find  the  sev- 
eral quantities  of  the  rest,  in  proportion  to  the  quantity  given. 

Rule. 
Take  the  differences  between  each  price,  andahe  mean  rate, 
and  place  them  alternately,  as  in  Case  1.  Then,  as  the  differenco 
standing  against  that  simple,  whose  quantity  is  given,  is  to  that 
quantity,  so  is  each  of  the  other  differences,  severally,  to  the  seve- 
ral quantities  required. 

Examples. 

1.  A  merchant  has  40lfe  of  tea,  at  Gs.  per  ft,  which  he  would  mix 
with  some  at  5s.  8d.  some  at  5s.  2(1.  and  some  at  4s.  6d.  :  How 
much  of  each  sort  must  he  take,  to  mix  with  the  40Ife,  that  he  may 
sell  the  mixture  at  5s.  5d.  per  ft  t 

ft 

li) 

10 

14 

14  stands  against  the  given  quantitv^ 
ft  s.  d. 

10  :  28y^j  at  4  6  ) 
10  :  28/^— -5  2)  per  ft 
14  :  40      —5  8) 

2.  A  farmer  being  determined  to  mix  20  bushels  of  oats,  at  GOc. 
per  bushel,  with  barley,  at  75c.  rye,  at  gl,  and  wheat,  at  ^1  25c. 
per  bushel  ;  I  demand  the  quantity  of  each,  which  must  be  mixed 
with  the  20  bnt;heis  of  oats,  that  the  whole  qijantity  may  be  worth 
90c.  per  busliel  ? 

•  Ans.  70  of  barley,  60  of  rye,  and  30  of  wheat,  (or  20  of  each.) 

3.  How  much  gold  of  16,  20  and  24  carats  line,  and  how  much 
illoy,  must  be  mixed  with  lOoz.  of  18  carats  line,  that  the  compo- 
Mtion  maybe  22  carats  fine. 

Ans.  lOoz.  of  16  carats  tine,  10  of  20,  170  of  24,  and  10  of  alloy, 


As  14  :  40 


:o2  ALIJGATiON  ALTERNATE. 

.      ALTERNATION  TOTAL.* 

CASE  IIL 

rVheji  the  rates  of  the  several  ingredients,  the  quantifi)  to  be  compound- 
ed, and  the  mean  rate  of  the  rvhole  mixture  are  given,  to  find  how 
much  of  each  sort  will  make  up  the  qnanlity. 

Rule. 
Place  the  difiercnccs  befweni  the  mean  ralp,  atid  (lie  several 
prices  alte^natel3^  as  in  (.'a?e  I  ;  tlien,  as  the  sorn  of  the  qnanti- 
ues,  or  differences  thus  deteitnined,  is  to  the  given  (|uantity,  or 
whole  composition  ;  so  is  the  difference  of  each  rate,  to  the  re- 
qiured  quantity  of  each  rate. 

Ea'amplfs. 

1.  Suppose  I  have  4  soi'ts  of  currant?,  at  TJd.  12d.  Vui\.  and  22d. 
per  Ye  ;  the  worst  will  not  sell,  and  the  he?t  are  too  dear;  I  there- 
fore conclude  to  mix  1^20Jfe  and  so  much  of  each  port  as  to  sell  them 
at  16(1.  per  Ife  ;   how  much  of  each  sort  must  1  take  ? 

d.        Ife  mm' 

r   8*<  6  '  fi)  :  3G  at     8d.T 

56c].  \  ^^JJ,^  ^^  oo  :    120  ::   ")  4  :  24  -  18d.  [P"'  ^ 
[22-^  8  [g  :  48  —  22d.  j     ' 

Sum— 20  120 

2.  A  goldsmith  has  several  sorts  of  gold  ;  viz.  of  15,  17,  20  antl 
22  carats  tine,  and  would  melt  toi^ether,  ofall  these  sorts,  so  much 
as  may  make  a  mass  of  40oz.  18  carats  fine  ;'  how  much  of  each 
sort  is  required  ? 

Ans.  ]6oz.  15"  carats  line,  8oz.  17,  4oz.  20,  and  12oz.  of  22  ca- 
rats line. 

*  I'o  llvi:^  Case  Lelong'-  iJuil  cnrionr,  qiiostion  concornin^  king'  Hiero's  crown. 

Tliero,  king  of  Sja-actise,  gave  orders  for  a  crown  \o  ttc?  made  entirely  of  pin  r 
;?;oid  ;  but  suspecting  the  "workmen  ]iad  debased  i I,  by  mixing  with  it  silver  oi 
copper,  he  recemmended  the  discovery  of  f he  fraud  Iq  th^  famous  Archiinedci, 
.uid  desired  to  know  the  exact  quantity  of  alloy  in  the  croivn,* 

Archimedes,  in  order  to  detect  the  imi>osition,  procured  two  other  itiasi^cr, 
one  of  pure  gold,  and  the  other  of  silver,  or  copper,  and  eadi  of  the  same  ^^eight 
with  the  former ;  and  by  j5utting  each  separately  hito  a, vessel  At//  of  Aval  or, 
the  quantity  of  water  expelled  by  tliem,  determined  their  .^pociiic  bulks  ;  from 
v/hich,  and  their  given  weights,  it  is  easier  to  determijic  the  quaiitities  of  gold 
and  alloy  in  the  crown  by  this  case  oi  Alhgatiou,  than  by  an  Algebr.iic  process. 

Suppose  the  weight  of  each  mass  to  have  been  [Ah.  the  Aveight  of  the  water 
expelled  by  the  alloy,  23oz.  by  the  gold,  13oz.  and  by  the  crown  lOoz.  tliat  is, 
!bat  their  specifick  bulks  Avere  as  23,  13,  "and  16  ;  then,  Avhat  Avcre  the  quanti- 
ties of  gold  and  alloy  respectively  in  the  crown  ? 

Here,  the  rates  of  the  simples  are  23  and  13,  and  of  the  compound  16,  whence, 
<  13      \7  of  gold  >  Antl  the 'sum  of  these  is  7-f-3=10,  Avhich  should  have 
(  23_  /3  of  alloy  \      been  but  5,  Avhence,  by  the  rule,     ' 
...     -       (  7  :  ?A\\).  of  gold  ) ,,      . 
^^•''■-   ^3:U]b.ufaHoyi^^^^"^^^''"^- 


16 


ALLIGATION  ALTERNATE.  vJ5r> 

o.  A  merchant  would  mix  4  sorts  of  wine,  of  several  prices,  viz. 
3t  75c.  gl  25c.  $1  60c.  and  $1  62-}c.  per  gallon  ;  of  these  he  wonld 
have  a  mixture  of  72  gallons,  worth  gl  37^0.  per  gallon  ;  what 
quantity  of  each  sort  must  he  have  ? 

Ans.  8  at  75c.  16  at  $1  25c.  40  at  gl  50c.  and  8  at  gl  62^Q. 
Or,  16  at  75c.  8  at  $1   25c.  8  at  $1  50c.  and  40  at  $  I   62ic. 

4.  How  many  gallons  of  water  of  no  value,  must  be  mixed  with 
wine,  at  4s.  per  gallon,  so  as  to  fill  a  vessel  of  80  gallons,  that  may 
be  afforded  at  2s.  9d.  per  gallon  ? 
Gal. 

Gal.  Gal.         Gal. 


i    0)15 
I  48^33 


33 

Sum  48 


A     <o     on        05  :  25  gallons  of  water.  }  . 
As  48  :  80  ::     <  ^c.      re:       n  r     •  i  Ans. 

(  33  :  55  gallons  ot  wme.    5 


CASE  IV* 

When  more  than  one  of  the  simples  are  limited. 

Rule. 
Find,  by  Alligation  Medial,  what  will  be  the  rate  of  a  mixture 
made  of  the  given  quantities  of  the  limited  simples  only;  then, 
consider  this  as  the  rate  of  a  limited  simple,  whose  quantity  is  the 
sum  of  the  first  given  limited  simples,  from  which,  and  the  rates  of 
the  unlimite<l  simples,  by  Case  11.  calculate  the  quantity. 

Examples. 

.  1.  How  much  wine,  at  80c.  and  at  87ic.  per  gallon,  must  be 
mixed  with  8  gallons  at  75c.  and  12  gallons  at  90c.  per  gallon,  that 
the  mixture  may  be  worth  82ic.  per  gallon  ? 

Limited  simples    \    ^  ^^"^"''  ^^  ^^^c.=$  6  > 

i.imiiea  simple^    ^  ^^  gallons,  at  90    =    10  80c.  $ 

20  16  80 

Gal.      $    c.      Gal.    c. 
As  20  :   16  80  ::   1   :  84  per  g-alion. 
Now,  having  found  the  rate  of  the  limited  simples,  the  question 
may  stand  thus  :  How  much    wine,  at  80c.  and  87ic.  per  gallon, 
must  be  mixed  with  20  gallons  at  84c.  per  gallon,  that  the  mixture 
may  be  worth  82Jc.  per  gallon  ? 


80   ^     m-5 

82A  /  84 


87i-^    2-1 


61  gallons,  at  80c. 
2  L 84 


As  n 


JgJ   ::20:   J^^  !!!!!!!!:i!  8^' £!!!!!!!" *  J  Answer. 


*  The  tbree  last  Cases  need  no  demonstration,  as  the  2d  and  3d  evidently 
result  from  the  fir>t,  an-l  the  last  from  AUi^alion  Medial,  and  the  second  Case  in 
Altcnrate. 

W  w 


\ 


POSITION. 

Proof. 
52  gallons  at  80c.     ~ 
CO       -     -       871      = 

8     -         -     75         = 
12       .     .       90        = 

^41   60c. 
17  50 
6 
10  80 

92     -         -     821      =       75  90 
2.  How  much  gold,  of  14  and  \6  carats  fine,  must  be  mixed  with 
6oz.  of  19,  and  12  of  22  carats  fine,  that  the  composition  may  be 
20  carats  fine  ?  Ans.  1  ^''joz.  of  each  sort. 


POSITION. 

POSITION  is  a  rule,  which,  by  false  or  supposed  numbers,  ta- 
ken at  pleasure,  discovers  the  true  ones  required.  It  is  divided 
into  two  parts  ;  single  and  double. 

SIJVGLE  POSITIOjY. 

Single  Position  teaches  to  resolve  those  questions,  whose  results 
are  proportional  to  their  suppositions :  such  are  those  which  re- 
quire the  multiplication  or  division  of  the  number  sought  by  any 
proposed  number ;  or  when  it  is  to  be  increased  or  diminished  by 
itself  a  certain  proposed  number  of  times. 

1.  Take  any  number,  and  perform  the  same  operations  with  it 
as  are  described  to  be  performed  in  the  question. 

2.  Then  say,  as  the  sum  of  the  errours  is  to  the  given  sum,  so  is 
the  supposed  number,  to  the  true  one  required. 

Proof.  Add  the  several  parts  of  the  sum  together,  and  if  it 
agrees  with  the  sum,  it  is  right. 

Examples. 

1.  A  school  master,  being  asked  how  many  scholars  be  had,  said, 
If  I  had  as  many  more  as  I  now  have,  three  quarters  as  many,  half 
as  many,  one  fourth  and  one  eighth  as  many,  I  should  then  have 
435:  Of  what  number  did  his  school  consist? 

*  The  operations  contained  in  the  question  heia*  performed  upon  tlie  answer 
or  number  to  be  fovmd,  will  give  the  result  contained  in  the  question.  The  same 
operations,  performed  on  any  other  number,  will  give  a  certain  result.  When 
the  results  are  proportional  to  their  supposed  numbers,  it  is  manifest  that  one 
result  must  be  to  the  result  in  the  question,  as  the  supposed  7iwnhcr  is  to  the  true 
(Hie  or  answer.  In  any  cases,  when  the  results  are  not  proportional  to  their  sup- 
positious, tlie  answer  cannot  be  found  by  tliis  rule. 


SINGLE  POSITION. 


Suppose  he  had  80 
As  many  =  80 
^  as  many  =  60 
A  as  many  =  40 
1  as  many  ~  20 
I-  as  many  —  10 

290 

As  290  :  435  ::  80 
80 

120 
120 
90 
60- 
30 
13 

29|0)3480|0(120  Ans. 
29 

58 
68 

0  435  Proof. 

2.  A  person  lent  his  friend  a  sum  of  money  unknown,  to  receive 
interest  for  the  same  at  6  per  cent,  per  annum,  simple  interest, 
and  at  the  end  of  12  years,  received  for  principal  and  interest 
J260  :  What  was  the  sum  lent  ?  Ans.  <J500. 

3.  A,  B  and  C  joined  their  stocks,  and  gained  ^353  12ic.  of 
which  A  took  up  a  certain  sura,  B  took  up  four  times  so  much  as  A, 
and  C,  five  times  so  much  as  B  :  What  share  of  the  gain  had  each  ? 

Cp4  12ic.  A's  share. 

Ans.    I    56  50"     B's  share. 

(  282  50       C's  share. 

4.  A,  B,  C  and  D  spent  353.  at  a  reckoning,  and,  being  a  little 
dipped,  they  agreed  that  A  should  pay  |,  B  i,  C  i,  and  D  I :  What 
did  each  pay  in  the  above  proportion  2  "  s.  d, 

fA,  13  4 
A  I  B,  10  0 
A"^-  1  C,    6  8 

[D,    6  0 

5.  A  certain  sum  of  money  is  to  be  divided  between  5  men,  ia 
such  a  manner  as  that  A  shall  have  i,  B  j,  C  y\,  D  ^rV*  and  E  the 
remainder,  which  is  £40  :  What  is  the  sum  ? 

Suppose  £200,  then  i4-i+y---f  ^i-=120. 

.200—120=80.     As  80  :  40  ::  200  :  100  Ans. 

6.  A  person,  after  spending  \  and  i  of  his  money,  had  £2G|  left : 
What  had  he  at  first?  "  Ans.  £1C0. 

7.  A  and  B,  talking  of  their  ages,  B  said  his  age  was  once  and 
an  half  the  age  of  A  ;  C  said  his  was  twice  and  one  tenth  the  age 
of  both,  and  that  the  sum  of  their  ages  was  93  :  What  was  the  age 
of  each  ?  Ans.  A's  is,  B's  18,  and  C's  63  years. 

8.  A  vessel  has  3  cocks,  A,  B  and  C  ;  A  can  fill  it  in  -^  an  hour, 
B  in  i  of  an  hour,  and  C  in  i  of  an  hour  :  In  what  lime  will  they 
all  fill  it  together?  Ans.  a  hour. 

9.  A  person  having  about  him  a  certain  number  of  dollars,  sai»l 
that  i,  1,  |,  and  i  of  Ihem  would  make  57  :  Pray,  how  many  had 
he?  Ans.  60. 

10.  A  Gentleman  bought  a  chaise,  horse  and  harness,  for  ^500 
the  horse  cost  }  more  than  the  harness,  and  the  chaise  ^  more  than 

l^the  horse  v What  wa*^  (ho  price  of  each? 

harness  ^127  6r.c.  9f^m. 
Jorse        159  57     4ff 


Ans.  ^.i 


Job  DOUBLE  POSITION. 

11.  A  and  B,  having  found  a  purse  of  money,  disputed,  who 
stiould  have  it :  A  said  that  i,  ^V  ^"^  irV  ^^  ^^  amounted  to  £35, 
and,  if  B  could  tell  him  how  much  was  in  it,  he  should  have  the 
whole,  otherwise  he  should  have  nothing  :  How  much  did  the  purse 
contain?  Ans.  £100. 

12.  A  gentleman  divided  his  fortune  among  his  sons  ;  to  A  he 
gave  g9,  as  often  as  to  B  ^5,  and  to  C  but  $o  as  often  as  to  B  g7, 
yet  C's  portion  came  to  ^1059  :  What  was  the  Whole  estate  ? 

Ans.  g7979  80c. 

13.  Seven  eighths  of  a  certain  number  exceeds  four  fifths  by  6  : 
What  is  that  number  ?  Ans.   80. 

14.  What  number  is  that,  wliich,  being  increased  by  |,  f  and  | 
of  itself,  the  sum  will  be  234|  ?  Ans.  90. 

DOUBLE  POSITIOX. 

Double  Position  teaches  to  resolve  questions  by  making  two 
suppositions  of  false  numbers. 

Those  questions,  in  which  the  results  are  not  proportional  to 
their  positions,  belong  to  this  rule  :  such  are  those,  in  which  the 
number  sought  is  increased  or  diminished  by  some  given  number, 
which  is  no  known  part  of  the  number  required. 

Rule.* 

1.  Take  any  two  convenient  numbers,  and  proceed  with  each 
according  to  the  conditions  of  the  question. 

2.  Place  the  result  or  errours  against  their  positions  or  suppos- 

Pos.      Err. 

30  12 

cd  numbers,  thus,      ^^       and  if  the  errour  be  too  great,  mark  it 

20-^^  6 
with  4-  ;  ^"tl  if  too  small  with  — . 

3.  Multiply  them  crosswise  ;  that  is,  the  first  position  by  the  last 
errour,  and  the  last  position  by  the  first  errour. 

*  The  rule  is  founded  on  this  supposition,  that  the  first  errour  is  to  the  se- 
cond, as  the  difference  between  the  true  and  first  supposed  number  is  to  the  dif- 
ference between  the  true  and  second  supposed  number  :  When  tliat  is  not  the 
case,  the  exact  answer  to  the  question  caimot  be  found  by  tliis  rule. 

That  the  rule  is  true  according  to  the  supposition  may  be  thus  demonstrated. 

Let  ^'2  and  B  be  any  two  numbers  produced  from  a  and  b  by  similar  opera- 
tions, it  is  required  to  find  the  number  from  which  JV  is  produced  by  a  like 
operation. 

Put  X  =  number  required,  and  let  jY—Ji—r^  and  .¥—/>=.<?.  Then  according 
to  the  supposition  on  wlpch  the  rule  is  founded,  r  ;  5  ::  x — a  :  x — i,  whence,  by 
multiplying  means  and  extremes,  rx — rb=zsx—sa  ;  and  by  transposition,  rx — 

rb — sa 

sx=^rb — sa ;  and  by  division,  x  = ==  number  sought ;  and  if  r  and  s  be 

r — s 
both  negative,  the  Theorem  is  the  same,  and  if  r  or  s  be  negative,  x  will  be 

rb-\-sa 
c^inal  to which  is  the  rule. 


DOUBLE  POSITION.  357 

4.  If  the  errours  be  alike,  that  is,  both  too  small  or  both  too 
gr^at,  divide  the  difference  of  the  products  by  the  difference  of 
the  errours,  and  the  quotient  will  be  the  answer. 

5.  If  the  errours  be  unlike  ;  that  is,  one  too  small,  and  the  other 
too  great,  divide  the  sum  of  the  products  by  the  sum  of  the  er- 
rours, and  the  quotient  will  be  the  answer. 

Note.  When  the  errours  are  the  same  in  quantity,  and  unlike 
in  quality,  half  the  sum  of  the  suppositions  is  the  number  sought. 

Examples. 
1.  A  lady  bought  damask  for  a  gown,  at  8s.  per  yard,  and  lining 
for  it  at  3s.  per  yard ;  the  gown  and  lining  contained  15  yards,  and 
the  price  of  the  whole  was  £3  10s. :  How  many  yards  were  there 
of  each  ? 

Suppose  6  yards  damask,  value  48s. 
Then  she  must  have  9  yards  lining,  value  27s. 

Sum  of  their  values  =  758. 
So  that  the  first  errour  is  5  too  much,  or  -f-     ^ 
Again,  suppose  she  had  4  yards  of  damask,  value  32s. 
Then  she  must  have  1 1  yards  of  lining,  value  o3s. 

Sum  of  their  values  ==  65s. 
.So  that  the  second  errour  is  5  too  little  or  —    5s. 
6^^5-1-  £>     s.  d. 

Then     >^  5  yards  at  8s.  =  2     0  0 

4^^^5—  10  yards  at  3s.  =  1    10  0 


20        30  3  10  0  proof. 

20 

Sum  of  errours  —  5-i- 5=^10)50 

Ans.    5yds.  damask,  and  15 — 5=  10yds.  lining. 
Or,  6+4—2=5  as  before. 

2.  A  and  B  have  the  same  income  ;  A  saves  }  of  his  ;  but  B, 
by  spending  £30  per  annum  more  than  A,  at  the  end  of  8  years 
finds  himself  £40  in  debt;  what  is  their  income,  and  what  doc- 
each  spend  per  annum  ? 

C    80_^120-[-     Ans.  Their  income  is  £200  per  ann. 
Suppose  {         V 

(  160-*-  ■*-   40-1-     Also,  A  spends  £175  and  B  £205  pci 
annum.     Then,  80 — 10=70  A's  expense  per  annum,  and   70-|-3(» 

=  100,  B's  expense  per  annum.  Then  100x8—80x8=160,  which 
should  have  been  40  ;  therefore,  IGO — 40=120  more  than  it  should 
be,  for  the  first  errour.  In  like  manner  proceed  for  the  second 
errour. 

3.  A  and  B  laid  out  equal  sums  of  money,  in  trade  :  A  gained  a 
sum  equal  to  i  of  his  stock,  and  B  lost  §225,  then  A's  money  was 
double  that  of  B  :  What  did'earh  lay  out  ?  Ans.  §600. 


%. 


.>5B  DOUBLE  POSITION. 

4.  A  labourer  was  hired  for  60  days  upon  this  condition,  thatlbi 
every  day  he  wrought,  he  should  receive  75c. ;  and  for  every  day 
he  was  idle,  should  forfeit  37ic.  ;  at  (he  expiration  of  the  time  he 
received  $18  :  How  many  days  did  he  work,  and  how  man^was 
he  idle  ?  Ans.  He  was  employed  36  days,  and  was  idle  24. 

5.  A  gentleman  has  two  horses  of  considerable  value,  and  a 
carriage  worth  £100  ;  now  if  the  first  horse  be  harnessed  in  it, 
he  and  the  carriage  together  will  be  triple  the  value  of  the  se- 
cond ;  but  if  the  second  be  put  in  they  will  be  7  times  the  value 
of  the  first :  What  is  the  value  of  each  horse  ? 

Ans.  One  £20  and  the  other  £40. 

6.  There  is  a  fish,  whose  head  is  10  feet  long  ;  his  tail  is  as  long 
as  his  head  and  half  the  length  of  his  body,  and  his  body  as  long 
as  the  head  and  tail :  What  is  the  whole  length  of  the  fish  ? 

Head=l0 
First,  suppose  the  body  20—.  ^  lO—      Tail  =30 

V  Body=40 

2d.  suppose  it  30-^^   5—  "       — 

Ans.  80  feet. 

7.  What  number  is  that,  which,  being  increased  by  its  i,  its  ], 
and  5  more,  will  be  doubled?  Ans.  20. 

8.  A  farmer,  having  driven  his  cattle  to  market,  received  for 
them  all  }J320,  being  paid  at  the  rate  of  $24  per  ox,  $16  per  cow, 
and  $6  per  calf;  there  were  as  many  oxen  as  cows,  and  4  times 
as  many  calves  as  cows  :  How  many  were  there  of  each  sort? 

Ans.  5  oxen,  6  cows,  and  20  calves. 

9.  A,  B,  and  C  built  a  ship,  which  cost  them  $5000,  of  which  A 
paid  a  certain  sum,  B  paid  $500  more  than  A,  and  C  $500  more 
than  both  ;  having  finished  her,  they  fixed  her  for  sea,  with  a  car- 
go worth  twice  the  value  of  the  ship :  The  outfits  and  charges  of 
the  voyage,  amounted  to  -}  of  the  ship ;  upon  the  return  of  which, 
they  found  their  clear  gain  to  be  §  off  of  the  vessel,  cargo  and 
expenses  :  Please  to  inform  me  what  the  ship  cost  them,  several- 
ly ;  what  share  each  had  in  her,  and  what,  upon  the  final  adjust- 
njcnt  of  their  accompts,  they  had  severally  gained  ? 

600.^^1500— 
Suppose  it  cost  A  ^l^ 

1000-^^500-f 
Ans.  A  owned  -^\  of  the  ship,  which  cost  him  $875,  and  his  share 
of  the  gain,   was  $1093  75c.  B  owned   ^1,  which  cost  $1375,  and 
his  gain  was  $1718  75c.  C  owned  ^J,  which  cost  $2750,  and  his 
gain  was  $3437  50c. 


PERMUTATIONS  AND  COMBINATIONS.  3o9 

PERMUTATIONS  AND  COMBINATIONS. 

THE  Premutation  of  Quantities  is  the  shewing^  how  many  dif- 
ferent ways  any  given  number  of  things  may  be  changed. 

This  is  also  called  variation,  alternation  or  changes  ;  and  the 
only  thing  to  be  regarded  here  is  the  order  they  stand  in  ;  for  no 
two  parcels  are  to  have  all  their  quantities  placed  in  the  same 
situation. 

The  Combination  of  quantities  is  the  shewing  how  often  a  less 
number  of  things  can  be  taken  out  of  a  greater,  and  combined 
together/  without  considering  their  places,  or  the  order  they 
stand  in. 

This  is  sometimes  called  election,  or  choice  ;  and  here  every 
parcel  must  be  different  from  all  the  rest,  and  no  two  are  to  have 
precisely  the  same  quantities,  or  things. 

The  Composition  of  Quantities  is  the  taking  of  a  given  number 
of  quantities  out  of  as  many  equal  rows  of  different  quantities,  one 
out  of  every  row,  and  combining  them  together. 

Here  no  regard  is  had  to  their  places  ;  and  it  differs  from  Com- 
bination only  as  that  admits  of  but  one  row  of  things. 

Combinations  of  the  same  form  are  those,  in  which  there  are 
the  same  number  of  quantities,  and  the  same  repetitions  ;  thus, 
a6cc,  bhad,  deef,  &c.  are  of  the  same  form  ;  but  abbcy  abbb,  aacc 
are  of  different  forms. 

Problem  I. 

To  find  the  number  of  permutations^  or  changes,  that  can  be  made  of 

any  given  number  of  things  all  different  from  each  other. 

KULE.* 

Multiply  all  the  terms  of  the  natural  series  of  numbers,  from  1 
up  to  the  given  number,  continually  together,  and  the  last  product 
wiil  be  the  answer  required. 

Examples. 

1.  Christ  church,  in  Boston,  has  8  bells:  How  many  changes 
may  be  rung  on  them  ?  1x2x3x4x5x6x7x8=40320  Ans. 

2.  Nine  gentlemen  n^et  at  an  inn,  and  were  so  pleased  with 
their  host,  and  with  each  other,  that  in  a  frolick,  they  agreed  to 
tarry  so  long  as  they,  together  with  their  host,  could  sit  every  day 
in  a  different  position  at  dinner:  Pray  how  long,  had  they  kept 
their  agreement,  would  their  frolick  have  lasted  ? 

Ans.  9941  If  4  years. 

3.  How  many  changes,  or  variations,  will  the  alphabet  admit  of? 

Ans.  G20448401 733239439360000. 

*=  The  reasoa  of  this  lule  may  hs  shewn  thus,  any  one  thing;  a  is  capable  of 
one  position  only,  us  a. 

Any  two  things  a  and  b  are  capable  of  two  variations  only  ;  as  o6,  ba ;  who've 
number  is  expressed  by  1  X2. 

If  there  be  three  things  a,  b  and  c ;  then  any  two  of  them,  leaving  out  tlie 
third,  w"l  have  1X2  variations  ;  and  consequently  when  the  third  is  taken  in, 
there  will  be  1 X2-X^  variations  ;  and  so  on,  a?  iuv  as  you  ploa*e. 


360  PERMUTATIONS  AND  COMBINATIONS. 

Problem  II. 

Jny  number  of  different  things  being  given^  to  find  how  many  changt^ 
can  be  made  out  of  them  by  taking  any  given  number  of  qudniitie? 
at  a  time. 

Rule.* 
Take  a  series  of  numbers,  beginning  at  the  number  of  thing? 
given,  and  decreasing  by  1,  as  many  terms  as  the  number  of  quan- 
tities to  be  taken  at  a  time  ;  the  product  of  all  the  terms  will  be 
the  answer  required. 

Examples. 
1 .  How  many  changes  may  be  rung  with  4  bells  out  of  8  ? 
8 
7 

66 
6 
-—        Or,  8X7X6X5  (=4  terms)  =1680  Ads. 

336 


1680 
2.  How  many  words  can  be  made  with  6  letters  of  the  alphabet^ 
admitting  a  number  of  consonants  may  make  a  word  ? 

24x23x22x21x20x19  (6  terms)  =96909120  Ans. 

Problem  III. 

'Any  number  of  things  being  given,  whereof  there  are  several  things 
of  one  sort,  several  of  another,  4'C.  to  find  how  many  changes  may 
be  made  out  of  them  all. 

RuLE.f 

1.  Take  the   series  1x2x3x4,  &c.  up  to  the  number  of  thing? 
given,  and  find  the  product  of  all  the  terms. 


*  This  Rule,  expressed  in  term?,  is  as  follows  ;  mXni — 1 X w — 2Xni — rj, &c. 
to  n  terms  ;  whence  m  =  number  of  things  given,  and  n  =■  quantities  to  be  ta- 
ken at  a  time. 

'4.mu-T>    1    •  1-    .  *i  1X2X3X4X5,&C.  tow. 

T  This  Rule  IS  expressed  m  terms  thus ; ~ — —7- 

^  lX2X3,&c.tojtfXiX2x3,Acc.to^,&:c- 

wlience  m  =  number  of  things  given,  p  =z  number  of  things  of  the  first  sort,  q 
=  number  of  things  of  the  second  sort,  &;c. 

Any  2  quantities,  a,  6,  both  different,  admit  of  2  clianges  ;  but  if  the  quanti- 
ties are  the  same,  or  ab  becomes  aa,  thei'e  will  be  only  one  alteration,  which  may 

1X2 

be  expressed  by =1 . 

'^  1X2 

Any  3  quantities,  «,  b,  c,  all  different  from  each  other,  admit  of  G  variations  -, 
but  if  the  quantities  are  all  alike,  or,  a  be  become  aaa,  then  the  6  variations 

1  X2X>3 

w^ill  be  reduced  to  1,  which  may  be  expressed  by —- -=:i.      Asain,  if  twf 

1X2X3  o      '    . 

quantities  out  of  three  are  alike,  or  abc  become  aac;  then  the  6  variations  will 

1X2x3 

be  reduced  to  these  3,rtar,  caa,  aca,' which  may  be  expressed  by  — •■-—  =  3. 

IX-*- 
and  so  of  any  greater  number. 


PERMUTATIONS  AND  COMBINATIONS.  361 

2.  Take  the  series  1x2x3x4,  &c.  up  to  the  number  of  the  giv- 
en things  of  the  first  sort,  and  the  series,  1x2x3x4,  &c.  up  to  the 
number  of  the  given  things  of  the  second  sort,  &lc. 

3.  Divide  the  product  of  all  the  terms  of  the  first  series  by  the 
joint  product  of  all  the  terms  of  the  remaining  ones,  and  the  quo- 
tient will  be  the  answer  required. 

Examples. 

1.  How  many  vatialions  may  be  made  of  the  letters  in  the  word 
Zaphnathpaaneah  ? 

1X2X3X4X5X6X7X8X9X10X11X12X13X14X16  (==  number  of 
letters  in  the  word)  =1307674368000. 

1X2X3X4X6  (=  number  of  as^  =  120 

1X2  (=  number  of  ps)  =      2 

1       (=  number  of  ^s)  =       1 

1x2x3  (==  nuiiiber  of  As)  =      6 

1X2  (=  number  of  ns)  =       2 

2X6X1X2X120=2880)1307674368000(454033600  Ans. 

2.  How  many  different  numbers  can  be  made  Qi  the  following 
figures,  1223334444?  Ans.  12600. 

Problem  IV. 

To  find  the  number  of  comhinations  of  any  given  number  of  things, 
all  different  from  one  another,  taken  any  given  number  at  a  time. 

Rule.* 

1.  Take  the  series  1,  2,  3,  4^  &,c.  up  to  the  number  to  be  taken 
at  a  time,  and  find  the  product  of  all  the  terms. 

2.  Take  a  series  of  as  many  terms,  decreasing  by  1,  from  the 
given  number,  out  of  which  the  election  is  to  be  made,  and  find 
the  product  of  all  the  terms- 

3.  Divide  the  last  product  by  the  former,  ard  the  quotient  will 
be  the  number  sought. 

Examples. 

1.  How  many  combinations  may  be  made  of  7  letters  out  of  12  ? 
1X2X3X4X5X6X7  (=  the  number  to  be  taken  at  a  time)=5040. 
12X11X10X9X8X7X6(=  same  number  from  I2)=3991680. 

5040)3091680(792  Ans. 

2.  How  many  combinations  can  be  made  of  6  letters  out  of  the 
24  letters  of  the  alphabet  ?  Ans.  134596. 

m      m — 1     m — 2    m — 3 

*  This  Rule,  expressed  algebraically,  is  —  x X X ,  fee.  to  n 

12  3  4 

terms ;  where  m  is  the  number  of  given  quantities,  and  n  those  to  be  taken  at  a 
time. 

Note.  In  any  given  number  of  quantities,  the  number  of  Combinations  in- 
vrcases  gradually  till  you  come  about  the  even  numbers,  and  then  gradually 
decreases.  If  the  number  of  quantities  be  ci'en,  half  the  number  of  places  will 
-]iew  the  greatest  number  of  Combinations,  that  can  be  made  of  those  quanti- 
Jes ;  but  if  odd,  then  those  two  numbers  which  are  the  middle,  and  whose  sum 
s  equal  to  the  givcu  number  of  quantities,  v.iU  show  the  :>;re.itest.  niimler  of 
•Jcmbiautions, 

X    T 


362 


TERMUTATIONS  AND  COMBINATIONS. 


1 1 -6 5 _^ 
*  '  1  1  34"* 


3.  A  general  was  asked  by  his  king-  what  reward  he  should  cori' 
fer  on  him  for  his  services  ;  the  general  only  required  a  penny  lor 
cverv  file,  of  10  men  in  a  file,  which  he  could  make  out  of  a  com- 
iiany  of  90  men  :   What  did  it  amount  to  ? 

Ans.  £23836022841 

4.  A  farmer  bargained  with  a  gentleman  for  a  dozeo  sheep,  (at 
2  dollars  per  head)  which  were  to  be  picked  out  of  2  dozen  ;  but 
being  long  in  choosing  them,  the  gentleman  told  him  that  if  he 
would  give»him  a  cent  for  every  different  dozen  which  might  be 
chosen  out  of  the  two  dozen,  he  should  have  the  whole,  to  which 
the  farmer  readily  agreed  :  Pray  what  did  they  cost  him  ? 

Ans.  §27041   56c. 

5.  How  many  locks,  whose  wards  differ,  may  be  unlocked  with 
a  key  of  6  several  wards  ? 

Ans.  63  :  6  of  which  may  have  one  single  ward,  15  double  wards, 
20  triple  wards,  15  four  wards,  6  five  wards,  and  1  lock  6  ward* 


Ward 

s.               Locks 

w 

ard 

s.                Locks. 

T 

'   61 

T 

"  5' 

2 

15 

2 

10 

- 

3 
4 

>    in  6  =  - 

20 
15 

► 

3 
4 

►   in  5  =  ' 

10 
5 

5 

L6J 

6 
.   1- 

Problem  V 

5 

1 

To  find  the  numlcr  of  combinations  of  any  given  number  of  things, 
by  taking  any  given  number  at  a  time  ;  in  which  there  are  several 
tilings  of  one  sort,  several  of  another,  <^'C. 

Rule. 

Fipd  the  number  of  different  fornix,  which  the  things,  to  be  ta- 
ken at  a  time,  will  admit  of,  in  the  following  manner : 

1.  Place  the  things  so  that  the  greatest  indices  may  be  first,  and 
the  r<»st  in  order. 

2.  Begin  with  the  first  letter  and  join  it  to  the  second,  third, 
fourth,  &c.  to  the  last. 

3.'  Join  the  second  letter  to  the  third,  fourth,  &c.  to  the  last  j 
and  so  on  till  they  are  all  done,  always  rejecting  such  combinations 
as  have  occurred  before  ;  and  this  will  give  the  combinations  of  all 
the  twos. 

4.  Join  the  first  letter  to  every  one  of  the  twos  ;  then  join  the 
second^  third,  &c.  as  before  ;  and  it  will  give  the  combinations  of 
all  the  threes. 

'5.  Proceed  in  the  same  manner  to  get  the  combinations  of  all 
the  fours,  fives,  &c.  and  you  will  at  last  get  all  the  several  ferms 
of  combination,  and  the  number  in  each  form. 

6.  Having  found  the  number  of  combinations  in  each  form,  add 
them  all  together,  and  the  sum  will  be  the  number  required. 


PERMUTATIONS  AND  COMBINATIONS. 


obo 


Example. 
Let  the  things  proposed   be  aaabbc :  It  is  required  to  find  the 
number  of  combinations  of  every  2,  of  every  3,  and  of  every  4 
of  these  quantities. 


Combinations  at  large. 

Forms. 

Combinations  in  each  form 

aa,aayab,ab,ac 

a2,62 

2 

aa,abyab,ac 

ab^acjjc 

3 

ab,ab,ac 

— 

bbM 

6=sum  of  the  twos. 

be 

a3 

1 

aaa,aabfaabyaac 

a^b,a^c,b^ 

a,62c4 

aab,aab^aac 

abc 

1 

abby(tbc 

.— 

bbc 

6=sum  of  the  threes. 

daab,aaab,aa(tc 

a^b.a^c 

o 

aabb^aabc 

aH^ 

1 

abbe 

a^bCfb^ac 

2 

Ans.  5  combinations  of  every  2  ; 
4  quantities. 


5=9um  of  the  fours, 
of  every  3,  and  6  of  every 


Problem  VI. 

To  find  the  changes  of  any  given  number  of  things^  taken  a  given, 
number  at  a  time ;  in  which  there  are  several  given  things  of  one 
sorty  several  of  another^  4*c. 

Rule. 

1.  Find  all  the  different  forms  of  combination  of  all  the  given 
things,  taken,  as  many  at  a  time,  as  in  the  question,  by  Problem  6. 

2.  Find  the  number  of  changes  in  any  form,    (by  Problem  3,) 
and  multiply  it  by  the  number  of  combinations  in  that  form. 

3.  Do  the  same  for  every  distinct  form,  and  the  sum   of  all  the 
products  will  give  the  whole  number  of  changes  required. 

Example. 
How  many  changes  can  be  m.ide  of  every  4  letters  out  of  these 

6,  aaabbc  1 


No.  of  forms.  Comb. 


a^b^a^c 
a"bcJ)"ac 


!! 


1X2X3X4=24 

1X2X3       =  6 
1X2X3X4=24 

1X2X1X2=  4 
1X2X3X4=24 


Changes. 


Qy 
I 


=  12 


364  MISCELLANEOUS  MATTERS 


(2X   4= 

I,  rix  6= 


Therefore,   <ix  6=  6 

:24 


38  =:  number  of  changes  required. 

Problem  VH. 

'i'b  find  the  (Compositions  of  any  number^  in  an  equal  number  of  sets^ 
the  things  being  all  different. 

Rule. 

Multiply  the  number  pf  things  in  every  set  continually  together, 
and  the  product  will  be  the  answer  required. 

Examples. 

1.  Suppose  there  are  five  companies,  each  consisting  of  9  men  ; 
it  is  required  to  find  how  many  ways  b  men  may  be  chosen,  one 
out  of  each  company  ? 

Multiply  9  into  itself  .continually,  as  many  times  as  there  are 
companies.    ,  9x9x9x9x9=59049  different  ways,  Ans. 

2.  How  many  changes  are  there  in  throwing  4  dice  ? 

As  a  die  has  6  sides,  multiply  6  into  itself  four  times  continually. 
6x6x6x6=1296  changes,  Ans. 

3.  Suppose  a  man  undertakes  to  throw  an  ace  at  one  throw  with 
4  dice,  what  is  the  probability  of  his  effecting  it  ? 

First,  6X6X6X6==1296  different  ways  with  and  without  the  ace. 
Then,  if  we  exclude  the  ace  side  of  the  die,  there  will  be  6  sides 
left ;  and  6x6X5X6=c625  ways  without  the  ace  ;  therefore  there 
are  1296 — 625=3^671  ways,  wherein  one  or  more  of  them  may 
turn  up  an  ace  ;  and  the  probability  that  he  will  do  it,  as  671  to 
625,  Ans. 

4.  In  how  many  ways  may.a  man,  a  woman  and  a  child  be  chos- 
en out  of  three  companies,  consisting  of  5  men,  7  women  and  ^ 
children?  Ans.  315. 


MISCELLANEOUS  MATTERS. 

A  short  method  of  reducing  a  Vidgar  Fraction,  into  its  equivalent 
Decimal,  by  Multiplication. 

Rule. 
Divide  unity  or  1  by  the  denominator,  till  the  remainder  is  a 
single  figure,  10,  100,  &c.  if  convenient,  then  multiply  the  whole 
quotient,  including  the  remainder  after  division,  by  the  remainder 
(whicli  is  now  the  numerator,  and  the  divisor,  the  denominator) 
and  annex  the  product  to  the  quptient,  then  multiply  the  quotient, 
thus  increased  by  tJie  last  numerator,  and  annex  the  product  to  the 


MISCELLANEOUS  MATTERS.  3&5 

increased  quotient ;  and  thus  it  may  be  reduced  to  what  exactness 
you  please.  But  if  the  numerator  of  the  given  fraction  exceed  1, 
you  must  iSnally  multiply  the  last  product  by  the  said  numerator. 

Examples. 

1.  Reduce  Jg^  to  its  equivalent  decimal. 

26)l-00(-03846^V 

78  This  multiplied  by  4  (the  numerator)  is  •15384i|=y\- 
- —  Which  annexed  to  the  quotient  -03846  is  -03846 15384-^^ 
220   And  -0384615384/^x8  and  annexed  to  the  last  product= 

208  '  -03846 lo3843076923076||,  &c, 

120 

104  • 

160 
166 

4 

2.  Reduce  -jfy. 

246)1'000000(-004065J^»^  and  -00406502^^X10  = -0040650^11 
and  this  annexed  to  the  quotient  is  -00406540650111,  and  this  mul- 
tiplied by  the  given  numerator,  5,  is  -02032703252^1^. 

For  any  number  of  pounds,  avoirdupois,  under  28,  multiply  the 
decimal  -00892857  by  the  given  number  of  pounds,  which  gener- 
ally gives  the  decimal  true  to  the  sixth  place. 

Jl  short  method  of  finding  the  duplicate^  triplicate^  ^c.  Hatio  of  any 
two  numbers,  whose  differmce  is  small,  compared  with  the  two  num- 
bers. 

FOR  THE  DUPLICATE  RATIO. 

Rule. 

Assume  two  numbers,  whose  difference  is  small  ;  subtract  half 
their  difference  from  the  least,  and  add  it  to  the  greatest,  and  the 
two  numbers,  thus  found,  will  be  in  the  same  proportion  nearly  as 
the  squares  of  the  assumed  numbers. 

Example. 

Let  the  assumed  numbers  be    10  and   11  ;    Then  II — 10=1. 
10— -5=9-5  and  llH-5=ll-5- 
Proof,  As  103  .  112  ..  9-5  .  1 1-5  nearly. 

FOR  A  TRIPLICATE  RATIO. 

Rule. 
Subtract  the  difference  of  the  assumed  numbers  from  the  least, 
and  add  it  to  the  greatest,  and  the  numbers,  thus  obtained,  will  be 
in  the  same  proportion  nearly  as  the  cubes  of  the  assumed  numbers. 


36d  J\IISCELLANEOtJS  MATTERS. 

Xet  the  numbers  be  164  and  165  :  Then  165—164—1.  164—- i 
=  163  and  165-f  1=-166. 

Proof,  As  1643  :  165^  ::  163  :  166  nearly. 

For  a  quadruplicate  proportion  subtract,  and  add  once  and  a  bah 
the  difference,  and  so  on,  for  each  higher  power,  increasing  the 
number  to  be  subtracted  and  added  by  '5. 

To  reduce  d  Ratio,  consisting  of  large  numbers,  to  its  least  terms, 
and  very  nearly  of  the  same  value, 

KULE. 

1.  Divide  the  greater  of  the  terms  by  the  less,  and  the  least  di- 
visor by  the  remainder,  and  so  on  continually,  till  nothing  remain, 
in  the  same  manner  as  we  get  the  greatest  common  measure  for 
reducing  a  vulgar  fraction :  This  will  give  a  number  of  ratios, 
from  which  we  can  choose  one,  that  will  suit  our  purpose. 

2.  Place  the  first  quotient  under  unit  for  the  first  ratio  ;  multi- 
ply that  by  the  next  quotient,  adding  nothing  to  the  numerator, 
and  1  to  the  product  of  the  denominator,  for  a  new  denominator, 
and  it  will  give  a  second  ratio,  nearer  than  the  first :  Then,  multi- 
ply the  last  ratio  by  the  next  quotient,  adding  the  preceding  ratio, 
and  so  on,  continually  till  you  have  gone  through. 

Examples. 
1.  Sir  Isaac  Newton  has  demonstrated,  in  his  Principia,  that  the 
velocity  of  a  comet,  moving  in  a  parabola,  is  to  that  of  a  planet, 

moving  in  a  circular  orb,  at  the  same  distance  from  the  sun,  as  y/  2 

to  1.     Let  this  be  taken  for  an  example. 

^"2=1-4142;  those  motions,  then,  are  as   1-4142  to  1  ;    or  as 

14142  to  10000  ? 

10000)14142(1  1 

1 0000  Then  -  =first  ratio . 

4442)10000(xJ  1x2+0=2 

8284  -  -=::zsecoad. 

— -  ^X  24  1=3 

1716)4142(-:  

3432  2X2+1=5 

-  -=:third.^ 

710)1716(2  3X2+1=7 

1420  " 

5X2+2=12 

296)710(2  -  — =fourt}i. 

592  7x2+3=17 

118)296(2  12X2+5=29       " 

236  -  -~=fi£ai,&e. 

17x2+7=41 

60)118(1 
60 

58)60(1 
58 

2)58(29 
58 

•  Thsr  late  Profesjror  Winthrop  chose  7  to  5  for  a  proportion. 


iMISCELLANEOUS  MATTERS. 


367 


2.  Geometers  have  found  the  proportion  of  the  circumference 
of  a  circle  to  its  diameter,  to  be  a?  3-1416  to  1  :  Let  this  ratio  be 
ii'educed. 
10000)31416(3  Then  ^=first  ratio. 

30000  

1X7+0     =     7 

-  — =secoud. 
3X7-f  1      =  22 

7X16+1   =113 

—  =third :  this  is  the 

22X16+3=355   ratio     generally 

made  use  of,  and 
issufficiently  ex- 
act for  very  nice 
calculations. 


1416)10000(7 
9912 


88)1416(1.6 
88 

636 
528 


8)88(11 
88 

3.  The  area  of  a  circle  is  to  its  circumscribing  square,  as  -7854 
to  I,  very  nearly :  Let  this  be  reduced. 


7854)10000(1 
7854 


2146)7854(3 
6438 


1416)2146(1 
1416 


1 
Then  -=first  ratio. 

I 1 

lX3-fO=  3 

-  -=second. 

1x3-1-1=  4 


730)1416(1 
730 


686)730(1 
686 


lUiird. 


=fourth. 


3X1-1-1=  4 

4x1  +  1=  5 

4x1+3=  7 

5x1+4=  0 

7x1+4=11 

'    — =fifth:    This 
9Xl-F5=14  is  very  exact, 
and  the  pro- 
portion   gen- 

erally  used. 

26  &c. 
Therefore,  as  14:  11  ::  the  square  of  the  diameter  of  a  circle  to 
its  area. 

To  estimate  the  Distance  of  Objects  on  level  ground,  or  at  sca^  having 
only  the  height  given. 


44)686(li 
44 

246 
220 


Hu 


LE. 


1.  To  the  earth's  diameter,  (viz.  42056462  feel,)  add  the  height 
•f  the  eye,  and  multiply  the  sum  by  that  height,  then  the  square  root 
of  the  product  is  the  distance,  at  which  an  object  on  the  surface  of 
the  earth  or  water,  can  be  seen  by  an  eye  so  elevated. 

2.  As  objects  are  seen  in  a  straight  line,  and  that  line  is  a  tangent 
to  the  earth's  surface  ;  therefore,  To  find  the  distame  of  two  elevat- 
ed objects,  rrhen  the  right  line  joining  them  touches  the  earth''s  surface 


:i6S  MISCELLANEOUS  QUESTIONS. 

between  those  objects,  (for  instance^  the  line  from  the  eye  of  the  ohserv- 
cr  to  the  distance  found  by  the  first  part  of  the  rule,  and  from  thence 
to  the  object ;)  work  for  each  object  separately,  and  th«  sum  of  tbe, 
square  roots  of  the  products  is  the  distauce  of  the  two  objects  from 
each  other. 

Example. 

How  far  may  a  mountaiu  be  seen  on  level  ground,  or  at  sea, 
which  is  a  mile  high,  supposing  the  eye  of  the  observer  elevated 
5  fe6t  above  the  surface  ? 

V42056462  -f  5   X   5=2-746  miles. 


-v/42056402-i-5280X5280=89-253  miles. 


Ans.  91-999  miles. 

To  estimate  the  Height  of  Objects  on  level  ground^  or  at  sea^  having 
only  the  distance  given, 

:     _-^.  Rule. 

1.  From  the  given  distance,  take  the  distance  which  the  elevation 
of  your  eye  above  the  surface  will  give,  found  by  the  last  problem. 

2.  Divide  the  square  of  the  remainder  in  feet  by  42056462  feet, 
and  the  quotient  will  be  the  height  required. 

Example. 

Being  on  my  return  from  a  foreign  voyage,  and  finding  by  my 
reckoning  I  was  about  5^  leagues  from  Boston  light  house,  it  being 
in  the  dusk  of  the  evening,  with  my  telescope  I  descried  the  lamp 
of  the  light  house  in  the  horizon,  at  which  time,  my  eye  was  ele- 
vated 6  feet  above  the  surface  of  the  water  :  Now,  supposing  my 
reckoning  to  be  true,  what  is  the  height  of  the  light  house  above 
the  ^ater?  '  

5^1eagues=16-5  miles;  then  16'5— ^42056462+6X6  =  13-943 
miles=^73619  feet  nearly,  and  73619x73619—42056462=129  feet 
nearly,  Ans. 

MISCELLANEOUS  QUESTIONS. 

1.  What  part  of  9d.  is  f  of  7d.  ? 

2       7      14  9     14      1X14     14 

5  ""^  T=T'*"*^  I'^T'^Wb  =45  ^"=- 

2.  What  number  is  that  from  which  ^  being  taken,  the  remain- 
der will  be  1  ?  Ans.  ||. 

3.  What  number  is  that,  to  which  iff  of  '/  of  ^^  be  added,  the 
total  will  be  1  ?  

3-      12        129      4644  1         4644       1 X 10955— 1  X4644_ 

T^  of  ~  of  ■;^=iyjj55i  and  j  —  10955=  1X10955         = 

6311 

Ans. 


[0955 
4.  What  number  is  that,  of  which  19y\  is  -f  ? 

19^^=2.vp  .  *»yor,    Aji  5  .  350  ..  7  .  2G'2  Am. 


MISCELLANEOUS  QUESTIONS:  369 

5.  In  an  orchard  of  fruit  trees,  }  of  them  bear  apples,  \  pears, 
^  plums,  60  of  them  peaches,  and  40  cherries  :  How  many  trees 
(Joes  the  orchard  contain  ?  ^04-40 

i4-i_UJ.=r:ii    and  i^ li=_i- •  therefore    as  -'-  •  ' 

::  if  :  1200  Ans.  ^ 

6.  A  person,  who  was  possessed  off  of  a  vessel,  sold  f  of  his 
interest  for  £375  :   What  was  the  ship  worth  at  that  rate  ? 

Ans.  £1500. 

7.  If  5  of  I  of  f  of  a  ship  be  worth  f  of  |  of  jf  of  the  cargo, 
valued  at  £1000  :   What  did  both  ship  and  cargo  cost? 

£837  12s.  l||d.  the  cost  of  the  ship;  and  £1837  12s.  Iffd- 
falue  of  the  ship  and  cargo,  Ans. 

8.  'J' wo  ships,  A  and  B,  sailed  from  a  certain  port  at  the  same 
time  ;  A  sailed  north  8  miles  an  honr,  and  B  east  G  miles  an  hour  : 
Required  by  an  easy  method,  to  find  their  distance  asunder  at 
every  hour's  end  ? 

10  miles  distant  in  1  hour,  and  10X2— 20  miles  in  2  hours,  &c.  Ans. 

9.  If  a  body  be  weighed  in  each  scale  of  a  balance,  whose  beam 
is  unequally  divided,  and  those  different  weights  of  the  body  be 
multiplied  together,  the  square  root  of  the  product  will  be  the  true 
weight  of  that  body. 

Suppose  the  weight  of  a  bar  of  silver,  in  one  scale,  to  be  lOoz. 
and  in  the  other  scale  12oz.  required  the  true  weight  of  the  bar  ? 

oz.  oz.pwt.gr. 

,      .  .        V12X10=10'95445+  =  10   19  2-1384-f-  Ans. 

10.  A  younger  brother  received  ^3125  92c.  which  was  just  -^\ 
of  his  elder  brother's  fortune  ;  and  5|  times  the  elder's  money 
was  1|  the  value  of  their  father's  estate  :  Pray,  what  was  their 
father  worth  ?  Ans.  gJ7281   87c.  2m.    , 

11.  A  gentleman  divided  his  fortune  among  his  sons,  giving  A 
£9  as  often  as  B£5,  and  to  C  but  £3  as  often  as  to  B£7,  and  yet 
€'s  dividend  was£l537|:   What  did  the  whole  estate  amount  to  ? 

Ans. £11583  8s.  lOd. 

12.  A  gentleman  left  his  son  a  fortune,  -V  of  which  he  spent  in 
.5  months,  f  of  f  of  the  remainder  lasted  him  9  months  longer, 
when  he  had  only  £537  left:  Pray,  w^hat  did  his  father  bequeath 
him? 

||=whole  legacy,  jf — re  — re  ^^^^  ^^  three'months,  then  |  of  ^ 

of  11=115     and    ') 1  fi5rr:i5.8i=r=_3J?_=  £^37  '  thprefore     a^*  JA. 

557  ::  1;  £2082   18s.  2f,-d.  Ans. 

13.  A  gay  young  fellow  soon  got  the  better  off  of  his  fortune  ; 
he  then  gave  £1500  for  a  commission,  and  his  profusion  continued 
till  he  had  but  £450  left,  which  he  found  to  be  just  j\  of  his  mo- 
ney after  he  had  purchased  his  commission  :  AVhat  was  his  fortune 
at'tirst?  Ans.  £3780. 

14.  A  merchant  begins  the  world  with  ^,5000,  and  finds  that  by 
his  distillery  he  clears  $5000  in  G  years  :  by  his  navigation  gSOOO 
in  7-1  years,  arxl  that  he  spends  in  gaming  <?.5000  in  ■'  yoars  :  How 
u;n<r  wiU  his  estate  last  ?  " 

Y  V 


370  MISCELLANEOUS  QUESTIONS. 


5000 


Asl  •    ^  -     S-  402 -r^^^'^  ^  travels. 


As  1666|—833i+666|  :  1  ::  5000  :  30  years,  Ans. 

15.  A  has  £100  of  B's  money  in  his  hands,  for  the  remittance  of 
which  B  allows  him  9  per  cent.  :  What  sum  must  he  remit,  to  dis- 
charge himself  of  the  £100?  Ans.  SlyVy- 

16.  Said  Harry  to  Edmund,  I  can  place  four  Is,  so  that,  when 
added,  they  shall  make  precisely  12  :   Can  you  do  so  too? 

Ans.  Hi. 

17.  A  and  B  are  on  o|>posite  sides  of  a  circular  field  268  poles 
about ;  they  begiti  to  go  round  it,  both  (ue  same  way,  at  the  same 
instant  of  time  ;  A  goes  22  rods  in  2  minutes,  and  B  34  rods  in  3 
minutes  :  How  many  times  will  they  go  round  the  field,  before  the 
swifler  overtakes  the  slower? 

min.  po.     mitj.     ])o. 

2  :  22  ^      -     (  1 1    A  goes  in  a  minute. 

3  :  34^  ■•  ^  •  ^  llAB     do.  do. 

therefore,  B  gains  11] — 1 1=^  of  a  pole  of  A  every  minute.     And, 
as  ipo.  :  Imin.  ;:  -f^po.  (=^half  round  the  field)  :  402min.  (=the 
time  in  which  B  will  overtake  A.)     Then, 
min.      ])o.  min.       uo. 

And,  VeV  —  ^^y  times  round  the  field,  A  travels  ; 
and  VeV^l"^  times  round  the  field,  B  travels. 

18.  If  15  men  can  perform  a  piece  of  work  in  1 1  days,  how  many 
men  will  accomplish  another  piece  of  work,  four  times  as  large  in 
a  fifth  part  of  the  time  ?  Ans.   300  men. 

19.  If  A  can  do  a  piece  of  work  alone  in  7  days,  and  B  in  12  ; 
let  them  both  go  about  it  together:  In  what  time  will  they  finish  it? 

Days. work.day  works,     work.  work.  work,  work.day.work.day. 

.     \  ?   S  Thpn  J-4--i-=.lll     As  15.  •  I"  1  •  4_8^  An<s 

/  12:  1  ::  1  :  -^  \  i  '  12     ai*  -^^  84  •  y  t  •  ^tq  ^"S* 

20.  A  and  B  together  can  build  a  boat  in  20  days  ;  with  the  as- 
sistance of  C  they  can  do  it  in  12:  In  what  time  would  C  do  it  by 
himself? 

D.  VV.  D.  W.  W.    W.     W.  W.  D.     W.     D. 

As  ]  To  :     ::     :  V  f  Then,  a -,L==-f-,&  as  8  :  l  ::  240  :  30  Ans. 

21.  A  can  do  a  piece  of  work  alone  in  13  days,  and  A  and  B  to 
gether  in  8  days  :   In  what  time  can  B  do  it  alone  ? 

Ans.  20|  days. 

22.  A,  Bj  and  C  can  complete  a  piece  of  work  in  12  days  ;  A 
can  do  it  alone  in  23  days,  and  B  in  37  days  :  In  what  time  can  C 
do  it  by  himself?  Ans.  77f|}  days. 

Another  question  ;  Four  persons  can  perform  a  certain  work  in 
the  following  manner,  viz.  A,  B,  and  C  can  do  it  in  6  days  ;  B,  C, 
and  D  in  7  days  ;  A,  C,  and  D  in  8  days,  and  .A,  B,  and  D  in  9  days  , 


MISCELLANEOUS  QUESTIONS.  371 

In  what  time  can  they  all  do  it  together,  and  in  what  time  can  eaclt 
one  do  it  alone  ? 

The  power  to  do  the  work  is  inversely  as  the  time  ;  whence. the 
power  of  A,  B,  and  C  will  be  i,  of  B,  C,  and  D  i,  of  A,  C,  and  D  |, 
and  of  A,  B  and  D  ^.  Hence  ^4-|-|-i+i=i-o24==5i-f>  's  the  power 
which  does  the  work  three  times,  for  each  agent  is  united  with 
others  three  times. 

Then  f^4X3=yJ|=5i5i  days,  the  time  in  which  all  together 
will  do  the  work. 

Then  rVra—o^jrVVz"^'^  power,  by  taking  A,  B,  and  C's  from 
the  sum  of  the  whole  power  to  do  the  work  once. 

Then  YaV  ^^y^^  ~  ^'^  ^''^^-  '"  ^^^^  ^^^^  way,  is  found  V^^d. 
=  A's  time.     Vt¥^-  —  ^'^  ^'*"^'  ^"^  Vri^-  ~  ^'^  'io^e,  Ans. 

23.  A  cistern,  for  water,  has  2  cocks  to  supply  it ;  by  the  first, 
it  may  be  filled  in  45  minutes,  and  by  the  second,  in  55  minutes; 
it  has  likewise  a  dischargmg  cock,  by  which  it  may,  when  full,  be 
emptied  in  30  minutes  ;  Now,  if  these  three  cocks  he  all  left  open, 
when  the  water  comes  in,  in  what  time  will  the  cistern  be  filled  ? 

Cist.  Hour.  Cist.  h.  m.  s. 
As  -4242  :    1   ::    1    :  2  21   20 1  Ans. 
Or,  by  vulgar  fractions,  more  accu- 
rately, 2h.  21m.  2543.  Ans. 


Gains  in  an  hour  '4242  of  a  cistern. 

24.  A  water  tub  holds  73  gallons ;  the  pipe,  which  conveys  the 
water  to  it,  usually  admits  7  gallons  in  5  minutes  ;  and  the  tap  dis- 
charges 20  gallons  in  17  minutes  :  Now,  suppof.ing  these  both  to 
be  carelessly  left  open,  and  the  water  to  be  turned  on  at  4  o'clock 
in  the  morning;  a  servant,  at  6,  finding  the  water  running,  puts  in 
the  tap  ;  in  what  time,  after  this  accident,  will  the  tub  be  full  ? 

Ans.  The  tub  will  be  full  at  32m.  58j  J-fs.  after  6. 

25.  A  has  a  chest  of  tea,  weighing  3icwt.  the  prime  cost  of  which 
is .£60:  Now,  allowing  interest  at  6  per  cent,  per  annum,  how  must 
he  rate  it  per  Ife  to  B,  so  that,  by  taking  his  note  of  hand,  payable 
at  G  months,  he  may  clear  §50  by  the  bargain  ? 

Interest  £2  5s.  Then  as  3icwt.  :  £60 -f  £15-1- £2  5s.  :;aife  : 
3s.  llffd.  Am. 

26.  Suppose  tltBt^merican  continental  debt  to  be  18  millions, 
what  annuity,  at  6  per  cent,  per  annum,  will  discharge  it  in  25 
years  ? 

By  Table  5,  of  annuities,  page  339,  -07823  is  the  annuity  which 
£1  will  purchase  in  25  years,  then,  -07823x18000000— 

£1408140    An? 
The  annual  interest  of  the  deht=  1080000 


3Iin. 

Cist. 

Min. 

Cist. 

45 

:   1   : 

:  GO   : 

;    1-3333 

5^ 

:    1   : 

:  GO  : 

;   10909 

2-4242 

30 

:   1    : 

:  GO  : 

;   2- 

Therefore,  there  must  be  a  sinking  fund  of  £328140  per  ann. 
27.  The  hour  and  minute  hands  of  a  watch  arc  exactly  together 
at  12  o'clock;  When  are  they  next  (ogether? 


B 


372  MISCELLANEOUS  QUESTIONS. 

The  velocities  of  the  two  hands  of  a  watch,  or  clock,  are  to  each 
other,  as  12  to  1  ;  therefore,  the  difference  of  velocities  is  12—1 

kc.  Ans. 


1  5  27fV) 

2  10  54/j-) 

3  16  2\j\) 


28.  A  hare  istarts  12  lods  before  a  hound;  but  is  not  perceived 
by  him  till  she  has  been  up  45  seconds  ;  she  scuds  away  at  the  rate 
of  10  miles  an  hour,  and  the  dog,  on  view,  makes  after  at  the  rate 
of  16  miles  an  hour  :  How  lono:  will  the  course  hold,  and  what  space 
will  be  run  over,  from  the  spot  wheie  the  dog  started? 

2288  feet  =  the  ground  run  over  by  the  dog.     9'7|sec.  Ans. 

29.  In  a  series  of  proportional  numbers,  the  first  is  4,  the  third 
12,  and  the  product  of  the  second  and  third  is  112-B:  What  is  the 
difference  of  the  second  and  fourth  ?  Ans.  18-8. 

3.0.  A  fellow  said  that  when  he  counted  his  luits;  two  by  two, 
three  by  three,  four  by  four,  five  by  live,  and  six  by  six,  there  was 
still  an  odd  one  ;  but  when  he  told  them  seven  by  seven,  they  came 
out  even  :   How  many  had  he  ? 

2X3X4X3X6=720,  and  720+T->7=103  even,  Ans.  721. 

31.  There  is  an  island  50  miles  in  circumference,  and  three  men 
start  together  to  travel  the  same  way  about  it :  A  goes  7  miles  per 
day,  B  8,  and  C  9:  When  will  they  all  come  together  again,  and 
how  far  will  each  travel  ? 

50x7-f50x8-f-50x9-~7+8-|-9'=50  days.     A  350  miles,  B  400,  and 
C  450,  Ans. 

32.  Suppose  A  leaves  Newburyport  at  6  o'clock  on  Monday 
morning,  and  travels  towards  Providence,  at  the  rate  of  4  miles 
per  hour  without  intermission  ;  and  that,  at  3  in  the  afternoon,  B 
sets  out  from  Providence  for  Newburyport,  and  travels  constantly 
at  the  rate  of  7  miles  an  hour:  Now  suppose  the  distance  between 
the  two  towns  to  be  90  miles  ;  whereabout  on  the  road  will  they 
meet  ? 

6-f  3=9  hours,  and  9x4=36  mile;i,  the  time  and  distance  A  had 
travelled  before  B  started.  Then  90—36=54  miles  remain  to  be 
travelled  by  both  ;  now,  as  both  togfether  lessen  the  distance  7+4 
=  11  miles  an  hour,  therefore  y\  of  54-|-36=55j\  miles  from 
Newburyport ;  which  is  near  Ames's,  at  Dedham. 

33.  If,  during  ebb  tide,  a  wherry  should  set  out  from  Haverhill 
to  come  down  the  river,  and  at  the  same  time,  another  should  set 
out  from  Newburyport,  to  go  up  the  river,  allowing  the  distance 
to  be  18  miles  ;  suppose  the  current  forwards  one  and  retards  the 
other  H  mile  per  hour;  the  boats  are  equally  laden,  the  rowers 
equally  good,  and,  in  the  common  way  of  working  in  still  water, 
would  proceed  at  the  rate  of  4  miles  per  hour  :  VVhere,  in  the  ri- 
ver will  the  two  boats  meet  ? 

Ans.  I2fm.  from  Haverhill,  and  5|m.  from  Newburyport. 
.34.  A  gentleman  making  his  addresses  in  a  lady's  family  who 
had  live  daughters  ;  she  told  him  that  their  father  had  made  a  will, 
"vhich  imported  that  the  first  four  of  the  girls'  fortunes   were,  to- 


MISCELLANEOUS  QUESTIONS.  373 

gether,  to  make  gSOOOO ;  the  last  four,  $66000;  the  three  last 
with  the  first,  g60000;  the  three  first  with  the  last,  56000 ;  and 
the  two  first  with  the  two  last,  $64000,  which,  if  he  would  unravel, 
and  make  it  appear,  what  each  was  to  have,  as  he  appeared  to 
have  a  partiality  for  Harriet,  her  third  daughter,  he  should  be 
welcome  to  her:  Pray,  what  was  Miss  Harriet's  fortune  ? 


A-fB+C-f-D        =30000 

B4.CH-D+E=66000 

A        +C-fD+E=60000 


A-fB        +D-|-E=64000 


296000 


Then,  296000—4  the  number  of 
combinations=74000  the  sum  of  their 
fortunes. 


A+B+C        +£=56000  ^  Then,  AH-B-1-C4-D4-E=74000 


And      A  +  B        +D-1-E=64000 


Ans.  Harriet's  fortune  =§10000 

35.  Three  persons  purchase  a  vessel  in  company,  towards  the 
payment  whereof  A  advanced  f ,  B  ^,  and  C,  $900:  What  did  A 
and  B  pay,  each,  and  what  part  of  the  vessel  had  C  ? 

^'^^  C's  part  of  the  vessel.  $2100  A  advanced.  $2250  B  ad^ 
vanced. 

36.  A  and  B  cleared,  by  an  adventure  at  sea,  45  guineas,  which 
was  £35  per  cent,  upon  the  money  advanced,  and  with  which  they 
agreed.to  purchase  a  genteel  horse  and  carriage,  whereof  they 
were  to  have  the  use  in  proportion  to  the  sums  adventured,  which 
was  found  to  be  11  to  A,  as  often  as  8  to  B  :  What  money  did  each 
adventure  ? 

As  JC35  :  100  ::  45  guineas  :  jei80=the  whole  adventure. 
-   As  11+8  •  180 -r^  '-^^^"^  4s.2i|d.  A's. 

37.  A,  B  and  C  are  to  share  £100  in  the  proportion  of  i,  ]  and 
}  respectively  ;  but  C  dying,  it  is  required  to  divide  the  whole 
gum  properlv,  between  the  other  two? 

£    9.    d. 

67     2  lOf  A's  share  in  all,  }  . 

42  17     14  B's  share  in  ail,  I  ^"^• 

Proof  100 

38.  A,  B  and  C  have  among  them  135  guineas  ;  A's-j-B's  are  to 
B's+C's,  as  5  to  7,  and  C's — B's  to  C'sH-B's  as  1  to  7  :  How  many 
had  each  ? 

A+B.  B+C. 
Suppose  A'sH-B's=50  ;  then,  as  5  :  7  ::  50  :  70  ;  as  7  :  1  ::  70  : 
iO=C's— B's;  then  70—10=60,  and  60-ir2=30=B's  ;  50—30= 
20=A's,  and  30+10=40=C's,  by  the  supposition  :  Now  20+30+ 
40=90,  which  should  have  been  135,  therefore, 

C  20  :  30=A's. 

As  90  :  135  ::  {30  :  45=B's. 

(40  :  60=C's. 

Snm=135  proof. 

39.  There  are  three  horses,  belonging  to  different  men,  employ- 
ed in  a  team  to  draw  a  l©«d  of  salt  froui  Newburyport  to  Boston, 


i^ 


374  MISCELLANEOUS  qUESTiOKS. 

for  £2  10s.  :  A  and  B  are  supposed  to  do  -j^  of  the  work  ;  A  and 
C  y5„  and  B  and  C  j\  of  it ;  they  are  to  be  paid  proportionally 
Can  vou  divide  it  as  it  should  be? 

Ans.  A's=:19|f|s. 
B's=:  9i||s. 
C's=21/^|s. 

Proof  60s.=the  sum. 

40.  I  would  put  20  hogsheads  of  London  beer  into  10  wine  pipes, 
and  desire  to  know  what  the  cask  must  contain,  which  will  receive 
the  difference,  231  solid  inches  being  the  wine  gallon,  and  282  that 
of  beer. 

Beer  hhd.  =  54  gall,  and    54x282x20=304560  solid  inches. 

Wine  pipe  =126  gall,  and  126x231x10=291060  solid  inches,  and 

304560—291060 

- — '■ ^2 =47f4  beer  gallons,  Ans. 

41.  Being  about  to  plant  5292  trees  equally  distant  in  rows,  the 
length  of  the  grove  is  to  be  three  tfmes  the  breadth  :  How  many 
of  the  shorter  rows  will  there  be  ? 

Ans.  viz.  i  of  the  trees  are  to  form  an  exact  square,  the  side 
whereof  being  42,  shews  how  many  come  into  a  short  row. 

42.  A  general,  disposing  his  army  into  a  square  battalion,  found 
he  had  231  over  and  above,  but  increasing  each  side  with  one  sol- 
dier, he  wanted  44  to  fill  up  the  square  :  How  many  men  did  his 
army  consist  of? 

231+44=275,  and  275^-^-2=137,  then  "137x137+231  =  19000 
Ans.     Proof,  138x138=19044. 

43.  1  want  the  length  of  a  shoar,  the  bottom  of  which,  beiwg  set 
9  feet  from  the  perpendicular  side  of  a  house,  will  support  a  weak 
place  in  the  wall,  22i  feet  from  the  ground  ? 

Ans.  24  feet,  2|  inches. 

44.  A  line  35  yards  long  will  exactly  reach  from  the  top  of  a 
fort,  standing  on  the  brink  of  a  river,  known  to  be  27  yards  broad, 
to  the  opposite  bank  :  What  is  the  height  of  the  wall  ? 

Ans.  22  yards,  9|  inches,  nearly. 

45.  Suppose  a  light  house  built  on  the  top  of  a  rock;  the  dis- 
tance between  the  place  of  observation  and  that  part  of  the  rock 
level  with  the  eye  620  yards  ;  the  distance  from  the  top  of  the  rock 
to  the  place  of  observation,  846  yards,  and  from  the  top  of  the 
light-house  900  yards  :  the  height  of  the  light-house  is  required  ? 

f~—  "-*~ '        f^ " 

V  900 X  900— 620x620— V  846x846—620x620=76 •77yds.  Ans. 

46.  Tfie  sum  and  difference  of  the  squares  of  two  numbers  given,  to 
find  those  numbers. 

Rule. 
From  the  sum  take  the  dilTerence,  and  half  the  remainder  is  the 
jquare  of  the  less,  %vhich,  taken  from  the  sum  of  the  squares;  wift 
give  the  square  of  the  greater. 


1 


MISCELLANEOUS  QUESTIONS.  o75 

A  and  B  have  between  them  a  number  of  guinea?,  which  are  to 
be  so  divided,  that  the  sum  of  their  squares  may  be  208,  and  the 
difference  of  their  squares  80;  supposing  A's  the  greaternumber, 
how  many  has  he  more  than  B  ? 

208—80—2=64  the  square  of  B's,  and  208—64=144  the  square 

of  A's;  therefore  ^144— -^64=4,  Ans. 

47.  Having  the  sum  of  two  numbers ^  and  the  sum  of  their  squares 

given,  to  find  those  numbers. 

Rule. 

From  the  square  of  their  sum  take  the  sum  of  their  squares  : 
then  from  the  sum  of  their  squares  take  this  remainder,  and  the 
square  root  of  the  difference  will  be  the  difference  of  the  two  num- 
bers. To  half  their  sum  add  their  difference,  and  the  sum  will  be 
the  greater.  From  half  the  sum  take  half  their  difference,  and 
the  remainder  will  be  the  less. 

A  and  B  have  50  guineas  between  them,  which  are  to  be  so  di- 
vided, as  that  the  sum  of  the  squares  of  the  two  numbers  shall  be 
1300 :  How  many  had  each,  supposing  A  to  have  the  greater  num- 
ber ?  

30x50—1300=1200  ;  then,  ^1 300— 1200=10  difference. 

Now  50—2+ 10-^2=30= A's.    And  50-t^— 10~-2=20=B's,  Ans. 

48.  Having  the  difference  of  two  numbers^  and  the  sum  of  their  squares 

given,  to  find  those  numbers. 

Rule. 

From  the  sum  of  their  squares  take  the  square  of  their  differ- 
ence :  to  the  sum  of  the  squares  add  the  remainder,  and  the  square 
root  of  thisj  sum  will  be  the  sum  of  the  required  numbers  ;  then, 
with  the  half  sum  and  half  difference  proceed  as  in  the  last  ques- 
tion. 

A  number  of  guineas  are  to  be  divided  between  A  and  B,  in  such 
a  manner  that  A  may  have  50  more  than  B,  and  that  the  sum  of  the 
squares  of  the  respective  shares  may  be  12500:  What  number  had 
each  ? 

12500— 50x50— 10000,  and  >/l2500-f-10000=150=sum  of  their 

shares.     Then,  Tbii^Z-\Eo^^=\00  A's  ;  and  150-r-2---'50^i= 
50  B's,  Ans. 

49.  Having  the  sum  of  the  squares  of  tm'o  numbers,  and  the  square  of 

their  half  sum  given,  to  find  thoss  numbers. 

Rule. 
From  the  sum  of  the  squares  take  twice  tbe  square  of  the  haU 
sum,  and  the  square  root  of  half  the  remainder  will  be  their  halt' 

^liffercnuo.  with  which  and  the  half  sum  proceei'  as  before  dirt^cted. 


:m  MISCELLANEOUS  qUESTIONS. 

« 

Let  tlie  sum  of  the  squares  of  two  numbers  be  3161,  and  Ihe 
square  of  their  half  gum  1560-25  :   Required  those  numbers  ? 

-3161—1560  25x2^40  5  40  5-t-2--20-25  and  v^20^5=:4-6=i 
difference, and  -v/1560-25=:39-5=i  sum  ;  then,  39  5+4-5—44  the 
greater,  and  39*5— 4  6=35  the  less,  Ans. 

50. — 1.  If  the  quantity  of  matter^  (or  weights)  of  any  two  bodies,  put 
in  motion,  be  equal,  ihe  force  by  which  they  are  moved  will  be 
in  .'proportion  to  their  velocities,  or  swiftness  of  motion. 

2.  If  the  velocities  of  these  bodies  be  equal,  their  forces  will  be  di- 
rectly as  the  quantities  of  matter  contained  in  them,  that  is,  as 
their  weights- 

3.  If  both  the  quantities  of  matter,  and  the  velocities  be  unequal,  the 
forces  with  which  the  bodies  are  moved,  will  be  in  a  proportion 

compounded  of  their  quantities  of  matter  and  velocities. 

Suppose  the  battering  ram  of  Vespasian  weighed  60000ife  ;  that 
it  was  moved  at  the  rate  of  24  feet  in  one  second,  and  that  this 
was  sufficient  to  demolish  the  walls  of  Jerusalem  :  With  what  ve- 
locity must  a  cannon  ball,  which  weighs  42lfe  be  moved,  to  do  the 
same  execution  ? 

The  velocity  of  the  ram  being  24,  and  the  weight  of  the  ball  42, 
compounded,  will  make  a  fraction— f|— 4,  and  4x60000=34285f 
feet  in  a  second,  Ans. 

61.  A  body  weighing  30ife  is  impelled  by  such  a  force  as  to  send 
it  20  rods  in  a  second  :  VVitli  what  velocity  would  a  body  weighing 
12lfe  move,  if  it  were  impelled  by  the  same  force  ? 

30X20 
— Y77— =50  rods  in  a  second,  Ans 

OF  GRAVITY. 

52,  The  gravity  of  bodies  above  the  surface  of  the  earth  decreases  in 
a  duplicate  ratio  (or  as  the  squares  of  their  distances)  in  semidiam- 
ciers  of  the  earth,  from  the  eartlis  centre. 

Supposing  a  body  to  weigh  400ife  at  2000  mile  ■»»iibove  the  earth's 
surface  :  What  would  it  weigh  at  the  surface,  estimating  the  earth's 
semidiameter  at  4000  miles  ? 

From  the  centre  to  the  given  height  being  IJ-  semidiameters  ; 
multiply  the  square  of  1|  by  the  weight,  and  the  f)roduct  will  be 
the  answer.  l-5xl'5X400=900ife  Ans. 

53.  If  a  body  weigh  900ife  at  the  surface  of  the  earth,  what  will 
it  weigh  at  2000  miles  above  the  surface  ? 

This  being  the  reverse  of  the  last,  therefore,  H- -5= 1-5  and  900 
-^l-5Xl'5'=400ifs  Ans. 

54.  A  certain  body  on  the  surface  of  the  earth,  weighs  ISOlfe  -. 
JIovv  high  must  it  be  carried  to  weigh  but  20115? 

v^l 80-^20=3,  Ans.  3  semidiameters  from  the  earth's  centre, 
that  is  8000  above  its  surface. 


I 


MISCELLANEOUS  QUESTIONS.  3T7 

bb.  To  what  height  must  a  ball  be  raised  to  lose  half  its  weight  ? 
As  1  :  398206X3982-06  ::  2  :  31713603-6872,  and  -v/3 17 13603-6872 
—5631-48  :  and  5631-48— 3982'06=1649-42  miles,  Ans. 

56.  Jit  ZDhat  distance  from  the  earth  zn'Oiild  a  balloon  be  suspended  be- 
tween the  earth  and  moon  ? 

Rule, 
As  the  Slim  of  the  square  roots  of  their  quantities  of  matter  is  to 
the  distance  of  their  centres,  so  is  the  square  root  of  the  quantity 
of  matter  in  the  earth,  to  the  distance  from  the  earth's  centre. 

The  proportional  quantity  of  matter  in  the  earth  being  to  that  in 
the  moon  as  4124  to  1:  and  the  distance  of  their  centres  240000-|- 
398206  +  1090:  therefore, as>/41-24+\/l:240000-f-3982U6-f  1090 
::  ^/41  24  :  212051-49.  And  21205149— 3982  06— 208069-43 
miles  from  the  earth's  surface,  Ans. 

57. — 1.  //  the  diameters  of  two  globes  be  equal,  and  their  densities 
different^  the  weight  of  a  body  on  their  surfaces  will  be  as  their 
densities. 

2.  If  their  densities  be  equals  and  diameters  diff'erent,  the  weight 
will  be  as  their  diameters. 

3.  If  their  diameters  and  densities  be  both  different,  the  weight  will 
be  as  the  product  of  their  diameters  and  densities. 

If  a  stone  weigh  lOOife  at  the  surface  of  the  earth,  required  its 
weight  at  the  surfaces  of  the  sun  and  the  several  planets,  whose 
densities  are  known  respectively  ? 

Sun.  Jupiter.     Saturn.      Earth.    Moon. 

TheiT  densities  100  78  5  36  392-5       464 

^  Diameters  in  >      883217  58.  89170-81.  79042-35.  7964-12     2180 
Eng.  miles,  y 


^        As  7964512x392-5  :  100:; 


883217-58X100    :  2825 •461b.  at  the  Sun. 

89170-8}  x7»-5  :  220-411b.  at  Jupiter. 

79042-35'x3d      :     91061b.  at  Saturn, 

L     2180      X464    :     32-35lb.  at  the  Moon. 


what  will  be  the  height  of  a  tide  raised  by  the  earth  on  the  sur- 
face of  the  moon  under  similar  circumstances? 

The  attraction  of  one  of  those  bodies  on  the  other's  surface  i** 
directly  as  its  quantity  of  matter,  and  inversely  as  its  diameter  ; 
therefore,  as  2180x2180x2180x464  :  5  ::  7964  x  7%4x7964x392v']^ 
■  206-22  directly.  And  as  2180  :  206-22  ::  7964  :  56448  inverse- 
ly,  Ans. 

OF  THE  FALL  OF  BODIES. 
59.  Heavy  bodies  near  the  surface  of  the  earth,  fall  one  foot  the 
iirst  quarter  of  a  second  ;  three  feet  the  second  quarter  ;  five  feet 
in  the  third,  and  seven  feet  in  the  fourth  quarter  ^  that  is,  16  ieet 
In  the  first  second.* 

■'-  Tiic  exact  velocil'*'  ;u  rr(n(o  i«  IG  ]  ia  ^];p  ^omnd  •  hv^  iv  ^hp  p'l  f^  ivjll  he, 


i78  MISCELLANEOUS  QUESTIONS. 

The  velocities,  acquired  by  bodies  in  falling,  are  in  proportion 
to  the  squares  of  the  times  in  which  they  fall ;  for  instance,  Let 
go  three  bullets  together  ;  stop  the  first  at  one  second,  and  it  will 
have  fallen  16  feet.  Stop  the  next  at  the  end  of  the  second  second, 
and  it  will  have  fallen  (2x2=  1)  four  times  16,  or  64  feet ;  and  stop 
the  last  at  the  end  of  the  third  second,  and  the  distance  fallen  will 
be  (3x3=9)  nine  times  16,  or  144  feet,  and  so  on. 

Or,  which  is  the  same,  the  space  fallen  through  (in  feet)  is  al- 
ways equal  to  the  square  of  the  time  in  4ths  of  a  second. 

Or,  by  multiplying  16  feet  by  so  many  of  the  odd  numbers,  be- 
ginning at  unity,  as  there  are  seconds  in  any  given  time  ;  viz.  by  I 
for  the  first  second,  by  3  for  the  second,  by  5  for  the  third,  and  so 
on,  these  several  products  will  give  the  spaces  fallen  through,  in 
each  of  the  several  seconds,  and  their  sum  will  be  the  whole  dis- 
tance fallen. 

The  velocity  given,  to  find  the  space  fallen  through* 

KULE. 

1.  The  square  root  of  the  feet,  in  the  space  fallen  through,  will 
ever  be  equal  to  one  eighth  of  the  velocity  acquired  at  the  end  of 
the  fall  ;  therefore, 

2.  Divide  the  velocity  by  8,  and  the  square  of  the  quotient  will 
be  the  distance  fallen  through,  to  acquire  that  velocity. 

Suppose  the  velocity  of  a  cannon  ball  to  be  about  }  of  a  mile,  or 
660  feet  per  second :  From  what  height  must  a  body  fall,  to  ac- 
quire the  same  velocity  per  second  ? 

660—8=82-5  and  82-6x82'6=6806i  feet,=lj\\  mile,  Ans. 

60.  Tlie  time  given,  to  find  the  space  fallen  through. 
KuLE. 

1.  The  square  root  of  the  feet,  in  the  space  fallen  through,  wrll 
ever  be  equal  to  four  times  the  number  of  seconds  the  body  has 
been  falling;  therefore, 

2.  Multiply  the  time  by  4,  and  the  square  of  the  products  will  be 
the  space  fallen  through  in  the  given  time. 

How  many  feet  will  a  body  fall  in  5  seconds  ? 

5X4=20,  and  20x20=400  feet,  Ans. 

61.  A  bullet  is  dropped  from  the  top  of  a  building,  and  found  tc» 
reach  the  ground  in  If  seconds  :     Required  its  height? 

1  -75x4=7,  and  7x7=49  feet,  Ans.  Or,  If =7qrs.  and  7x7=49. 
Or,  l-75xl-75xl6=49feet,  Ans. 

62.  What  is  the  difference  between  the  depths  of  two  wells,  into 
each  of  which  should  a  stone  be  droj)})ed  in  the  same  instant,  one 
would  reach  the  bottom  in  5  seconds,  and  the  other  in  3  ? 

5x4=20,  and  20x20=400  feet, 
3xl  =  12,.and  12x12=144  feet. 

Ans.  256  fett. 

63.  Ascending  bodies  are  rclardcd  in  the  same  ratio  that  de- 
scending bodies  are  accelerated  ;  therefore,  if  a  ball,  discharged 
from  a  guu,  return  to  Ihc  earth  in  12  seconds  :  Uo»v  high  did  it  as- 
cend ? 


MISCELLANEOUS  QUESTIONS.  37H 

The  ball  being  half  of  the  time,  or  6  seconds,  in  its  ascentjthere- 
fore,  6x4=24,  and  24x24=576  feet,  Ans. 

64.  The  velocity  per  second  given,  to  Jind  the  lime.. 

RULK. 

1.  Four  times  the  number  of  seconds,  in  which  a  body  has  been 
falling,  is  equal  to  one  eighth  of  thn  velocity,  in  feet,  per  second, 
acquired  at  the  end  of  the  fail ;  therefore, 

2.  Divide  the  given  velocity  by  8,  and  one  fourth  part  of  the 
quotient  will  be  the  answer. 

How  long  must  a  bullet  be  falling,  to  acquire  a  velocity  of  160 
feet  per  second  ?  160-f-8— 20,  and  20—4=5  seconds,  Ans. 

65.  The  space  through  which  a  body  has  fallen,  given,  to  find  the  time- 
it  has  been  falling. 

RuLK. 

1.  Four  times  the  number  of  seconds,  in  which  the  body  has 
been  falling,  will  ever  be  equal  to  the  square  root  of  the  space, 
in  feet,  through  which  it  has  fallen  ;  therefore, 

2.  Divide  the  square  root  of  the  space  fallen  through  by  4,  and 
the  quotient  will  be  the  time,  in  which  it  was  falling. 

In  how  manv  seconds  will  a  bullet  fall  tlirough  a  space  of  10125 
feet  ?     Vl0125=100-6,  and  100  6^-4=25-15  seconds=25"  9"'  AbS. 

66.  In  what  time  will  a  musket  ball,  dropped  from  the  top  of  a 
steeple,  484  feet  high,  come  to  the  ground  ? 

V'484=22,  and  22-^4=51  seconds,  Ans. 

67.   To  fin(Jl  the  velocity,  per  second,  with  which  a  heavy  body  will 
begin  to  descend,  at  any  distance  from  the  eartli's  surface. 

Rule. 

As  the  square  of  the  earth's  semidiameter  is  to  16  feet,  so  is 
the  square  of  any  other  distance  from  the  earth's  centre,  inverse- 
ly, to  the  velocity  with  which  it  begins  to  descend  per  second. 

With  what  velocity,  per  second,  will  an  iron  ball  begin  to  de- 
scend if  raised  3000  miles  above  the  earth's  surface  ? 
As  4000X4000  :  16  ::  4000+3000x4000+3000  :  5.22449  feet,  Ans, 

68.   How  high  must  a  ball  be  raised  above  (he  earth's  surface, 
to  begin  to  descend  with  a  velocity  of  522449  feet  per  second  ? 
As  16  :  4000X4000  ::  5-22449  :  49000000,  and  V49000()6o"=7000. 
Wherefore,  7000—4000=3000  miles,  Ans. 

69.  To  find  the  mean  velocity  of  a  falling  body. 

Rule. 
Divide  the  space  fallen  through  by  the  number  of  seconds  it  was 
falling,  and  the  quotient  will  be  the  mean  velocity. 

A  musket  ball  dropped  from  the  top  of  a  steeple  484  feet  high 
in  51  seconds  :    Required  its  mean  velocity  ? 

484-~5-5— 88  feet  per  seroi^l.  Am. 


380  MISCELLANEOUS  QUESTIONS. 

70.  To  find  the  velocity  acquired  by  a  falling  body^  per  second,  (or 
by  a  stream  of  water,  having  the  perpendicular  descent  given)  at 
the  end  of  any  given  period  of  time. 

Rule. 

1.  The  velocity  acquired  at  the  end  of  any  period  is  equal  to 
twice  the  mean  velocity,  with  which  it  passed  during  that  period. 

Or,  2.  Multiply  the  perpendicular  space  fallen  through  by  64, 
and  the  square  root  of  the  product  is  the  velocity  required. 

If  a  ball  fall  through  a  space  of  484  feet  in  5i  seconds,  with  what 
velocity  will  it  strike  ? 
By  the  former  part  of  the  rule.  By  the  latter  part,  with- 

484-4-5*5—88,  and  out  regarding  the  time. 

88x2=176,  Ans.  ^484x64=176,  Ans. 

71.  There  is  a  sluice,  or  flume,  one  end  of  which  is  2i  feet  low- 
er than  the  other  :   What  is  the  velocity  of  the  stream  per  second  ? 
2-6x64=160,  and  ^160=12-649  feet,  Ana. 

72.   The  velocity,  zcith  which  a  falling  body  strikes,  given,  to  find  the 
space  fallen  through. 

Rule. 
Divide  the  square  of  the  velocity  by  64,  and  the  quotient  will 
be  the  height  required. 

If  a  ball  strike  the  ground  with  a  velocity  of  56  feet  per  second, 

from  what  height  did  it  fall  ?  

5t)X56-r.64=49  feet,  Ans. 
73.  The  mean  velocity  of  a  fluid,  or  stream,   is  1^-649  feet  per 
second  :  What  is  the  perpendicular  fall  of  the  stream  ? 

12-649xl2-649-~64=2i  feet,  Ans. 

74.  The  weight  of  a  body,  and  the  space  fallen  through,  given,  to  find 

the  force  with  which  it  will  strike. 

Rule. 

The  momentum,  or  force,  with  which  a  falling  body  stiikes,  is 
equal  to  its  weight  multiplied  by  its  velocity  ;  tiierefore,  find  the 
velocity,  by  Problem  70,  and  multiply  it  by  the  weight,  which  ^vill 
produce  the  force  required. 

If  the  rammer,  used  for  driving  ihe  piles  of  Charlestown  bridge, 
weighed  2^  tons,  or  4500lfe  and  fell  through  a  space  of  U)  feet,  with 
what  force  did  it  strike  the  pile  ■ 

V'10x64=:±25  3=velocity,  and  26-3x4500=^  1138501fe  momen- 
tum, Answer. 

75.  The  weight  and  momentum,  or  /striking  force,  given,  to  find  the 

space  fallen  through. 

Rule. 
Divide  the  momentum   by  the  weight,  and  Ibe  quotient  will  be 
Uie  velocity  ;  then  divide  the  square  of  the  velocity  by  64,  and  the 
quotient  will  be  the  ?pace  fallen  through. 


MISCELLANEOUS  QUESTIONS.  381 

If  the  aforementioned  rammer  weighed  4500ife  and  struck  with  a 
force  of  1138501fe  :  From  what  heig^ht  did  it  fall  ? 

n3850~4500=25-3,  and  25^3x2 5~3~6 4 —  -[0  feet,  Ans. 

76.  If  it  were  required  to  know  with  what  quantity  of  motion, 
momentum  or  force,  a  fluid,  moving  with  a  given  velocity,  strikes 
upon  a  fixed  obstacle, 

Rule. 

By  Problem  72  find  the  fall,  which  will  produce  the  given  velo- 
city ;  multiply  that  height  by  62-51fe  Avoird.  for  clean' river  water, 
by  63}fe  for  dirty  water,  and  by  64  for  sea  water. 

Suppose  a  stream  of  clear  water  to  move  at  the  rate  of  5  feet 
per  second,  and  to  meet  with  a  fixed  obstacle  (or  bulk  head)  15 
feet  wide  and  4  feet  high  :  What  is  the  momentary,  instantaneous 
pressure  of  the  stream  ? 

5X5~~64=ff  and  26-T-64="39  of  a  foot,  for  the  perpendicular 
fall  of  the  water.  Now  62-5X'39=24-375lfe  the  pressure  upon 
each  square  foot,  which,  multiplied  by  60,  (the  number  of  square 
feet  in  the  obstacle)  gives  14625lfe  going  with  the  given  velocity 
of  5  feet  per  second  ;   therefore,  1462'5x  5=7312-5ife  Ans.* 

77.  The  velocity  of  water,  spouting  through  a  sluice,  or  aper- 
ture in  a  reservoir,  or  bulk  head,  is  the  same  that  a  body  would 
acquire  by  falling  through  a  perpendicular  space  equal  to  that  be- 
tween the  top  of  the  water  in  the  reservoir  and  the  aperture. 

What  is  the  velocity  of  water  issuing  from  a  head  of  5  feet  deep  ? 
By  Problem  70lh  64x5=320,  and  ^/:32(T=18  feet  nearly,  Ans. 

78.  If  the  velocity  of  a  stream  issuing  through  the  bulk  head  of 
a  mill,  be  16  feet  per  second,  what  head  of  water  is  there. 

l6xl6"-^64=4  feet,  Ans. 

79.  The  quantity  of  water  discharged  from  a  hole  in  a  vessel, 
is  as  the  square  root  of  the  height  of  water  above  the  aperture. 

A  miller  has  a  head  of  water  4  feet  above  the  sluice  :  How  high 
must  the  water  be  raised  above  the  opening,  so  that  half  as  much 
again  water  may  be  discharged  from  the  sluice  in  the  same  time  ? 

-v/4=2,  and  half  as  much  again  as  2,  is  2-1-1^=3,  for  the  square 
root  of  the  required  depth  ;  therefore,  3x3=9  feet  high,  Ans. 

OF  PENDULUMS. 

80.  The  time  of  a  vibration,  in  a  cycloid,  is  to  the  time  of  a 
heavy  body's  de.-cent  through  half  its  length,  as  the  circumference 
of  a  circle  to  its  diameter,  that  is,  as  3- 14 16  to  1  :  therefore,  (as 
a  body  descends  freely,  by  gravity,  through  about  193'5  inches  in 
the  first  second)  to  find  the  length  of  a  pendulum  vibrating  se- 
conds. 

Rule. 
As  31416x31416  :  1x1  ::  193  5  :  196  inches,  the  half  length, 
and  19-6X2=39-2  inches,  the  length. 

*  Water  being  a  yielding  substance,  loses  two  thipTs  of  it?  power  in  produc- 
m2:  effects. 


382  MISCELLANEOUS  (QUESTIONS. 

81.  To  find  the  length  of  a  pendulum^  that  "will  swing  any  given  time. 

Rule. 

Multiply  the  square  of  the  second?  in  any  given  lime  by  39-2 
and  the  product  will  be  the  length  required,  in  inches. 

Required  the  lengths  of  several  pendulums,  which  will  respec- 
tively swing  i  seconds,  |  seconds,  seconds,  niinutes  and  hours  ? 

•25X-25X39-2=24o  inches  for  ^  seconds.  •5X-5X39-2=9-8 
inches  for  A  seconds,  ixl  x39'2=39-2  inches  for  seconds,  as 
above;  60X60x39-2=the  inches  in  2  miles  and  1200  feet,  for 
minutes;  and  1  hour=3C00  seconds,  therefore  3600X3600X392 
:=^the  inches  in  8018  miles  and  96  feet,  for  hours,  Ans. 

82.  What  is  the  difference  between  the  length  of  a  pendulum, 
which  vibrates  half  seconds  and  one  which  swings  three  seconds  ? 

3X3X39^— •5X-5X39"^=:343  inches=28/^  feet,  Ans. 

83.  To  find  the  time  which  a  pendulum  of  any  given  length  will  swing. 

Rule. 

Divide  the  given  length  by  39-2,  and  the  quotient  will  be  the 
square  of  the  time  in  seconds. 

Or,  as  6-261  (the  square  root  of  39-2)  is  to  the  square  root  of  the 
given  length,  so  is  1  second  to  the  time  of  1  oscillation  :  that  is,  di- 
vide the  square  root  of  the  given  length  by  6*261,  and  the  quo- 
tient will  be  the  time  of  one  vibration  of  that  pendulum. 

How  often  will  a  pendulum  of  9  8  inches  vibrate  in  a  second  ? 

By  the  former  part  of  the  rule,  9-8^39-2=*25  of  a  second,  and 
^/'25=^■5  of  a  second,  the  time  of  one  vibration,  that  is,  it  vibrates 
balfseconds,  or  60~-5=120  limes  in  a  minute. 

By  the  latter  part.  V9^=3-13,  and  v'39-2=6-261,  therefore, 
3;13-r6-26]=  o  of  a  second. 

84.  I  observed,  that  while  a  stone  was  falling  from  a  precipice, 
a  string,  (with  a  bullet  at  the  end)  which  measured  25  inches,  (to 
the  middle  of  the  ball,)  had  made  5  vibrations  :  What  was  the 
iieight  of  the  precipice  ?         

25-r39-2=:-6377+,  and  >/6377=-7986— of  a  second,  the  time  of 
one  vibration,  and  -7986X5=1  seconds,  nearly,  the  time  of  the 
ftone's  descent ;  then  4X4=16,  and  16x16=256  feet,  Ans. 

c5.   To  find  the  true  depth  of  a  welly  by  dropping  a  stone  into  ity  also 
the  time  of  the  stone's  descent  and  of  (he  sound's  ascent. 

^  Rule. 

1.  Take  a  line  of  any  length,  and  by  the  last  Problem  find  the 
ti'me  from  the  dropping  of  the  stone  till  you  hear  it  strike  the  bot- 
tom. 

2.  Multiply  73088  (=16X4X1 142;  1 142  feet  being  the  distance, 
which  sound  moves  jn  a  second)  by  the  number  of  seconds  till  you 
hear  the  stone  strike  the  bottom. 

3.  To  this  product  add  1304164  (=the  square  of  1142)  and  from 
the  square  root  of  the  sum  take  1142. 

4.  Divide  the  square  of  the  remainder  by  64  (=16X4)  and  the 
quotient  will  be  the  depth  of  the  well  in  feet. 


MISCELLANEOUS  QUESTIONS.  383 

o.  Divide  the  depth  by  1142,  and  the  quotient  will  be  the  time 
of  the  sound's  ascent,  which,  being  taken  from  the  whole  time,  wili 
leave  the  time  of  the  stone's  descent  in  seconds. 

Suppose  I  drop  a  stone  into  a  well,  and  a  string  with  a  plummet, 
tvhich  measured  to  the  mitdle  of  the  ball,  25  inches,  made  5  vi- 
brations before  I  heard  the  stone  strike  the  bottom  :  Required  the 
depth,  time  of  the  stone's  descent,  and  of  the  sound's  ascent : 

25-^39-2=-6377,  and  v''6377=-7986,  and  -7986X5—4  seconds 


to  the  hearing  of  it  strike  ;  then,  -v/73088X44. 1304164— 1142  = 
121-53;     and    121  53X  121-53-f-64=230-77  feet,  the    depth,    and 
23077-r-1142=-2  of  a  second,  the  time  of  the  sound's  ascent,  and 
4 — •2=3-8  seconds,  the  time  of  the  stone's  descent. 
OF  THE  LEVER  OR  STEELYARD. 

86.  It  is  a  principle  in  mechanicks,  that  the  power  is  to  the 
weight,  as  the  velocity  of  the  weight,  to  the  velocity  of  the  power. 
Therefore,  to  find  what  weight  may  be  raised  or  balanced  by  any 
given  power,  say  ; 

As  the  distance  between  the  body  to  be  raised  or  balanced,  and 
the  fulcrum  or  prop,  is  to  the  distance  between  the  prop  and  the 
point  where  the  power  is  applied  ;  so  is  the  power  to  the  weight 
which  it  will  balance. 

If  a  man,  weighing  160ife,  rest  on  the  end  of  a  lever  10  feet 
long,  what  weight  will  he  balance  on  the  other  end,  supposing 
the  prop  one  foot  from  the  weight  ? 

The  distance  between  the  weight  and  prop  being  1  foot,  the 
distance  from  the  prop  to  the  power  is  10 — 1=9  feet  ;  there- 
fore, as   1  foot  :  9  feet  ::   160ife  :   1440ife,  Ans. 

87.  If  a  weight  of  1440lfe  were  to  be  raised  with  a  lever  10  feet 
long,  and  the  prop  fixed  one  foot  t>om  the  weight,  what  power  or 
weight,  applied  to  the  other  end  of  the  lever  would  balance  it  ? 

As  9  :   1   ::   1440  :   1601fe,  Ans. 

88.  If  a  weight  of  1440Ife  be  placed  1  foot  from  the  prop,  at 
what  distance  from  the  prop  must  a  power  of  160lfe  be  applied,  to 
balance  it?  As   160  :   1440  ::   1   :  9  feet,  Ans. 

89.  At  what  distance  from  a  weight  of  1440ife,  must  a  prop  be 
placed,  so  as  that  a  power  of  1601fe,  applied  9  feet  from  the  prop 
may  balance  it?  As   1440  :   160  :;  9  :   1   foot,  Ans. 

90.  In  giving  directions  for  making  a  chaise,  the  length  of  the 
shafts  between  the  axletree  and  backhand,  being  settled  at  9  feet,  a 
dispute  arose  whereabout  on  the  shafts  the  centre  of  the  body 
should  be  fixed.  The  chaise  maker  advised  to  place  it  30  inches 
before  the  axletree  ;  others  supposed  20  inches  would  be  a  sulTi- 
cient  incumbrance  for  the  horse  :  Now  supposing  two  passengers  to 
weigh  3cwt.  and  the  body  of  the  chaise  |cwt.  more  :  VVhat  will  the 
beast  in  both  these  cases  bear,  more  than  his  harness  ? 

Weight  of  the  chaise  and  passengers  3|cwt.--420ifc,  and  9  f^et- 
tOB  inches.  ,  „  In.         ft 

*"•       ''*        i  u)  •  nG?-> 

Then,  as  1C8   :   420  ;:    >o     '      iti|  J  Ans. 


384  MISCELLANEOUS  QUESTIONS. 

OF  THE  WHEEL  AND  AXLE. 
01.  The  proportion  for  ihe  wheel  and  axle  (in  which  ihe  power 
is  applied  to  the  circumference  of  the  wheel,  and  the  weight  is 
raised  by  a  rope,  which  coils  about  the  axle  as  the  wheel  turns 
round)  is,  as  the  diameter  of  the  axle  is  to  the  diameter  of  the  wheel, 
so  is  the  power  applied  to  the  wheel,  to  the  weight  suspended  by 
the  axle. 

A  mecbanick  would  make  a  windlass  in  such  a  manner,  as  that 
life  applied  to  the  wheel,  should  be  equal  to  lOlfe  suspended  from 
the  axle  ;  now,  supposing  the  axle  to  be  six  inches  diameter,  re- 
quired the  diameter  of  the  vvheel  ? 
11).       in.      lb.       in. 
As   10  :  6  ::   1    :  60  inversely,  the  diameter  required. 

92.  Suppose  the  diameter  of  the  wheel  to  be  CO  inches:  Re- 
quired the  diameter  of  the  axle,  so  as  that  life  on  the  wheel  may 
balance  lOlfe  on  the  axle  ? 

lb.      in.  lb.      in. 

Inversely,  as   1   :  60  ::   10  :  6  diamett.r  required. 

93.  Suppose  the  diameter  of  the  axle  6  inches,  and  that  of  the 
wheel  60  inches,  what  power  at  the  wheel  will  balance  lOIfe  at  the 
axle?  in.     lb.       in.     lb. 

Inversely,  6   :    10  ::  60  :    1    Ans. 

94.  Suppose  the  diameter  of  the  wheel  60  inches,  and  that  of  tbft 
axle  6  inches ;  what  weight  at  the  axle  will  balance  life  at  fhe  wheel  ? 

in.       lb.      in.       lb. 
Inversely,  as  60  :   1    ::  6  :   10  Ans. 
OF  THE  SCREW. 

95.  The  power  is  to  the  weight,  which  is  to  be  raised,  as  the 
distance  between  two  threads  of  the  screw,  is  to  the  circumference 
of  a  circle  described  by  the  power  applied  at  the  end  of  thelever. 

Rule. 

Find  the  circumference  of  the  circle  described  by  the  end  of  the 
lever;  then,  as  that  circumference  is  to  the  distance  between  the 
spiral  threads  of  the  screw ;  so  is  the  weight  to  be  raised,  to  the 
power  which  will  raise  it,  abating  the  friction,  which  is  not  propor- 
tional to  the  quantity  of  surface  ;  but  to  the  weight  of  the  iiicijrn- 
bent  part;  and,  at  ci  medium,  -}  part  of  the  effect  of  the  macliine 
is  destroyed  by  it,  sometimes  more  and  sometimes  less. 

There  is  a  screw,  whose  threads  are  an  inch  asunder  ;  the  level* 
by  which  it  is  turned  30  inches  long,  and  the  weight  to  be  raised  a 
ton,  or  22408i :  What  power  or  Ibrce  must  be  applied  to  the  end  of 
the  lever,  sufficient  to  turn  the  screw — that  is,  to  raise  the  weight. 

The  lever  being  the  semi<lian)eter  of  the  circle,  the  diameter  is 
60  inches  ;  then,  3-1 116X60=  180- 196  inches,  the  circumference  ; 

in.  in.  lb.  lb. 

Therefore,  as   188-496  :    1   ::  2240  :    1188,  Ans. 

96.  Let  the  lever  be  30  inches,  (the  circunjference  of  which  is 
fodnd  to  be  188-496)  the  threads  1  inch  asunder,  and  the  power 
n-88fc  :   Required  the  weight  to  be  raised  ? 

in.        in.  lb.  lb. 

As  1   :   188-496  ::   11-88  :  2240  nearly,  Ans. 


MISCELLANEOUS  QUESTIONS.  385 

97.  Let  the  weicfht  be  2240ife,  the  power  ll-88!fe,  and  the  lever 
30  inches  :   Required  the  distance  between  the  threads  ? 

lb.  lb.  in.  in. 

As  2240  :   11-88   ::  288-496   :   1  nearly,  Ans. 

98.  Let  the  power  be  11 -88115,  the  weight2240ife,  and  the  threads 
ao  inch  asunder,  to  find  the  length  of  the  lever. 

lb.  lb.  in.         in. 

As  11-88  :  2240  ::  1  :  188-5;  then,  as  355  :  1 13  ::  188-5  :  60 
inches  nearly,  the  diameter,  and  C0-r-2=30  inches,  Ans. 

99.  Suppose  one  of  those  meteors,  called  fire  balls,  to  move  par- 
allel to  the  earth's  surface,  and  50  miles  from  it,  at  the  rate  of  20 
miles  per  second  :  In  what  time  would  it  move  round  the  earth  ? 

The  Earth's  diameter  is  7964  English  miles ;  then,  7964  +  50x2  = 
8064  =  the  diameter  of  the  circle  described  by  the  ball.  Then, 
8064x3~1 4 16=25333-8624  miles^its  circumference,  and  25333-8624 
-f-20=1266-69312  seconds=21'  6"  41'"  35""  .13""' 55"  ""  12""",  Ans. 

100.  Sound,  uninterrupted,  moves  about  1142  feet  in  a  second 
How  long,  then,  after  firing  a  cannon  at  Newburyport,  before  It  will 
he  heard  at  Ipswich,  estimating  the  distance  at  10  miles  in  a  right 
line? 

10  miles=52800  feet,  and  52800-rl  142=46i^f  seconds,  Ans. 

101.  In  a  thunder  storm  I  observed  by  my  clock  that  it  was  6 
seconds  between  the  lightriing  and  thfinder  :  at  what  distance  was 
the  explosion?  11 42x6=6852feet=:li^i  miles,  Ans. 

102.  Tubes  may  be  made  of  gold,  weighing  not  more  than  at  the 
rate  of  y^^j^  of  a  grain  per  foot  :  What  v/ould  be  the  weight  of 
such  a  tube,  which  would  extend  across  the  Atlantick,  from  Boston 
to  London,  estimating  the  distance  at  1000  leagues? 

1000X3=3000  miles,  and  3000x5280=15840000  feet,  and 
15840000XToV5=9747/^gr.  or  rather,  life  8oz.  6pwt.  3-^^g£.  Ans- 

103.  The  mean  distances  of  the  Planets  from  the  Sun,  in  English 
miles,  are  as  follow:  viz.  Mercury  36686617:5  ;  V^enus  6855213583; 
Earth  94772980;  Mars  144404783-33;  Jupiter  49291253333  ;  Sat- 
urn 903957657  5  :  Now,  as  a  cannon  ball,  at  its  first  discharge,  flies 
about  a  mile  in  8  seconds,  and  sound  1142  feet  in  a  second  :  In 
what  time,  at  the  above  rate,  would  a  bullet  pass  from  the  Earth  to 
the  Sun  ?  and  sound  move  from  the  Sun  to  Saturn? 

94772980x8"=758183840=24years,15days,6hours,27min»ite3, 
20  seconds,  for  the  passage  of  the  ball.  And  903957657-5x5280~' 
4772896431600  feet,  and  477289643 1600-M142--n  132  years,  192 
days,  21  h.  42m.  21||^-s.  sound  passing  from  the  Sun  to  Saturn,  Ans. 

104.  Light  passes  from  the  Sun  to  the  Earth  in  8-2  minutes  :  In 
what  time  would  it  pass  from  the  sun  to  the  Geor'^ium  Sidus,  it 
being  1803930416-66  English  miles  ? 

As  '94772980  :  8-2  ::  1803930416  66  :  2h.  36m.  4"  50'",  Ans, 

105.  The  Sun's  diameter  is  883217-58  English  miles  ;  .Tupiter's 
is  89170-81;  Saturn's  7904235;  Georgium  35109;  Mercury's 
322248;    Venus'    7687-85;    Earth's    7964-12;    Mar6Ml89C9  ; 

A  3 


..u6  3fISCELLAiSE0US  qUESTIOIsS. 

and  the  Moon's  2180:    Requirerl  the  comparative  magnilutle  be 
tween  each  of  those  bodies  and  the  Earth  ? 


}i«a'2l7-58  X  88:52l7-5}^  X  883217-58 

89170-81  X 

89170-81  X 

89170-81 

-  9704i2-35X 

97042-35  X 

9704'2-35 

36109      X 

35109       X 

35109 

f964-12x7964-12x7964'12— 


[1363724 

1402-65 

rt> 

982 

1 

99-57 

(  3222-48  X  3222-48  X  .i222-4B=93- 12 
—^  ;  7687-85X7687-85X7687-85=  Ml 
^-    \  4189-69x4189-69x4189-69=  6-86 

'       2180X2180      X2180     =48-74 


7964- 12X7964- 12x7964-12—  <  IVon.^nC,  Von/^nf.  .Von.^nZ  ^.o^  >  tess 


x; 

N.*B.  The  above  diameters  and  mean  distances  in  English  miles 
answer  to  the  same  in  geographical  miles,  as  they  were  deduced 
from  observations  on  the  transits  of  Venus  over  the  Sun  in  1761 
and  1769. 

106.  Suppose  the  density  of  the  Moon  464,  and  that  of  the  Earth 
392-5  :  Required  the  proportion  between  the  quantity  of  matter  In 
the  Earth  and  in  that  of  the  Moon,  allowing  the  Earth's  diameter 
to  be  796412,  and  the  Moon's  2180  miles,  and  supposing  the  Earth 
a  complete  sphere,  which,  however,  it  is  not? 

7964-12X7964' 12X7964- 12X392-6 
There  is -g^-     X2180       X^O       X464    ^^^'^^    ^^"^^«    *^^ 
quantity  of  matter  in  the  Earth  that  there  is  in  the  Moon  ;  or,  the 
Earth's  weight  is  so  many  limes  that  of  the  Moon. 

107.  The  mean  diameter  of  the  Earth's  orbit,  (or  annual  path 
round  the  Sun)supposing  it  a  circle,  is  in  English  miles  190437141'7: 
Required  its  mean  niolion,  (or  the  space  through  which  it  moves 
in  its  orbit,)  per  miimte  ? 

19043714I'7X311i6— 598'n7324-36  miles  in  circumference; 
then, 

Days. 

As  366-25-:  598277324-36  ::  1'  :  1 137-49  miles,  Ans. 

N.  B.  The  Earth's  diurnal  motion  round  its  axis  is  ll^  miles  per 
minute,  at  the  equator. 

OF  THE  SPECIFICK  GRAVITIES  OF  BODIES. 

The  specifick  gravities  of  bodies  are  as  their  densities,  or  weights^ 
bulk  for  bulk  ;  thus,  p.  body  is  said  to  have  two  or  three  times  the 
speciiick  gravity  of  another,  when  it  contains  two  or  three  times 
as  much  matter  in  the  same  space. 

A  body,  immersed  in  a  fluid,  will  sink,  if  it  be  heavier  than  its 
bulk  of  the  Huid.  If  it  be  suspended  therein,  it  will  lose  so  much 
of  what  it  weighed  in  the  air,  as  its  bulk  of  the  fluid  weighs. 
Hence,  all  bodies  of  equal  bulk,  which  will  sink  in  fluids,  lose  equal 
weights  when  suspended  therein,  and  unequal  bodies  lose  in  pro- 
portion to  their  bulks. 

The  hydrostaiick  balance  differs  very  litde  from  a  common  bal- 
ance that  is  nicely  made  ;  only  it  has  a  hook  at  the  bottom  of  each 
scale,  on  which  small  weights  may  be  hung  by  horse  hairs,  so  that 
a  body  suspended  by  the  hair,  may  be  immersed  in  water  wi-thou' 
wetting  the  scales. 


MISCELLANEOUS  QUESTIONS.  33: 

How  to  find  the  Specifick  Gravities  of  Bodies. 

if  the  body,  thus  suspended  under  the  scale,  at  one  end  df  the 
balance,  be  first  counterpoised  in  air  by  weights  in  the  opposite 
scale,  and  then  imnnersed  in  water,  the  equilibrium  will  be  imme- 
diately  destroyed  ;  then,  if  as  much  weight  be  put  into  the  scale, 
to  which  the  body  is  suspentled,  as  will  restore  the  equilibrium, 
(without  altering  the  weights  in  the  opposite  scale)  that  weight, 
which  restores  the  equilibrium,  will  be  equal  to  a  quantity  of  water 
as  big  as  the  immersed  body  ;  and  if  the  weight  of  the  body  in 
air  be  divided  by  what  it  loses  in  water,  the  quotient  will  shew  how 
much  that  body  is  heavier  than  its  bulk  of  water.  Thus,  if  a  guin- 
ea suspended  in  air,  be  counterbalanced  by  129  grains  in  the  oppo- 
site scale,  and  then,  upon  being  immersed  in  water,  it  becomes  so 
much  lighter  as  to  require  7}  grains  to  be  put  into  the  scale  over 
it,  to  restore  the  equilibrium,  it  shews  that  a  quantity  of  water,  of 
equal  bulk  with  the  guinea,  weighs  7  25  grains  ;  by  which  divide 
129  (the  weight  of  the»guinea  in  air)  and  the  quotient  will  be 
17-793  ;  which  shews  that  the  guinea  is  17*793  tiines  as  heavy  as 
its  bulk  of  water. 

Thus  may  any  piece  of  gold  be  tried,  by  weighing  it  first  in  air, 
and  then  in  water;  and  if,  upon  dividing  the  weight  in  air  by  the 
loss  in  water,  the  quotient  comes  out  17*793,  the  gold  is  good  :  If 
the  quotient  be  18,  or  between  18  and  19,  the  gold  is  very  fine  : 
but  if  it  be  less  than  17,  the  gold  is  too  much  alloyed  by  being  mix- 
ed with  some  other  metal. 

;  If  silver  be  tried  in  this  manner  and  found  to  be  11  times  as  hea- 
vy as  water,  it  is  very  fine  :  It  it  be  10^  times  as  heavy,  it  is  stand- 
ard ;  but  if  it  be  of  any  less  weight  compared  with  water,  it  is  mix- 
ed with  some  lighter  metal,  such  as  tin,  k,c. 

If  a  piece  of  brass,  glass,  lead,  or  silver,  be  immersed  and  sus- 
pended in  different  sorts  of  fluids,  the  different  losses  of  weight 
therein  will  shew  how  much  heavier  it  is  than  its  bulk  of  the  fluid  ; 
that  fluid  being  lightest,  in  which  the  immersed  body  loses  least  of 
its  aerial  weight. 

Common  clear  water,  for  common  uses,  is  generally  made  a 
standard  for  comparing  bodies  by,  whose  gravity  may  be  represent- 
ed by  unity,  or  1,  or,  in  case  great  accuaracy  be  required,  by  1*000, 
where  3  cyphers  are  annexed  to  give  room  to  express  the  ratios 
of  other  gravities  in  larger  numbeis  in  the  table.  In  doiuj-^  this 
there  is  a  twofold  advantage  ;  the  first  is,  that,  by  this  mean  the 
soecifick  gravities  of  bodies  may  be  expressed  to  a  much  greater 
degree  of  accuracy.  The  second  is,  that  the  numbers  of  the  Ta- 
ble, considered  as  whole  numbers,  do  also  express  the  ounces 
Avoirdupois  contained  in  a  cubick  foot  of  every  sort  of  matter  there- 
in specified;  because  a  cubick  foot  of  common  water,  is  found  by 
experiment  to  weigh  very  nearly  1000  ounces  Avoirdupoi'5,  or  62i. 
pounds. 


388 


MISCELLANEOUS  QUESTIONS. 


A  TABLE 

OE  THE  SPECIFICK  GRAVMTIES  OF  SEVERAL  SOLID  AND  FLUID  BODIES ;  WHERE 
THE  SECOND  COLUMN  CONTAINS  THEIR  ABSOLUTE  WEIGHT,  AND  THE  THIRD 
THEIR  RELATIVE  WEIGHT,  IN  AVOIRDUPOIS  OUNCES. 


A  Cubick  Foot  of 


jAbso,  I  Rela. 
\   wt.   I    wt. 


A  Cubick  Foot  of      i 


Abso.l   Rela. 


I  wt.  I     wt. 


Platina  rendered  mal-  ) 
leable  and  hammered  ^ 
Very  fine  Gold  -  - 
Standard  G  old  -  - 
Guinea  Gold  -  -  _ 
Moidore  Gold  -  -  - 
Quicksilver  -     -     -     - 

Lead 

Fine  Silver         -     _     _ 
Standard  Silver    -     - 
Rose  Copper       -     -     - 
Copper  -     -     -     -     - 
Plate  Brass    -     -     -     - 

Steel 

Cast  Brass     -     -     -     - 

Iron  ------ 

Block  Tin     -     -     -     - 

Cast  Iron     -     -     -     - 

Lead  Ore       -     -     -     - 

Copper  Ore      -     -     - 
t)iamond       -     -     -     - 

Crystal  Glass   -     -     - 
White  Marble    -     -     - 
Black  Marble  -     -     - 
Rock  Crystal     -     -     - 
Gi-een  Glass       -     -     _ 
Clear  Glass      -     ^     - 


20170  20-170 


fFlint  -  - 
c^     Pavinof  - 
St^^^i  Cornelian 

(^Free  -  - 


19637 

18888 

17793 

17140 

13600 

11325 

11087 

10535 

9000 

8843 

8000 

785 

7850 

7645 

7321 

7135 

6800 

3776 

3400 

3150 

2707 

2704 

2658 

2620 

2600 

2582 

2570 

2568 

2352 


19-637 

18-888 

17-793 

17-140 

13-600 

11-325 

11-087 

10-535 

9-000 

8-843 

8-000 

7-852 

7-850 

7-645 

7-321 

7-135 

6-800 

3-775 

3-400 

3-150 

2-707 

2-704 

2-658 

2-620 

2-600 

2-582 

2-570 

2-56 

2-352 


Brick 

Liver  Sulphur  -  - 

Nitre 

Alabaster  -     -     -     - 
Dry  Ivory    -     -     -     - 
Brimstone       -     -     - 
Solid  subs,  of  Gun  Pow, 
Alum      ----- 
Ebony        -     -     -     - 
Human  Blood  -     -     - 
Amber       -     -     -     - 
Cow's  Milk       -     -     - 
Sea  Water      -     -     - 
Pure  Water      -     -     - 
Red  Wine      -     -     - 
Oil  of  Amber    -     -     - 
Ptoof  Spirits  -     -     - 
Dry  Oak      -     -     -     - 
Olive  Oil   -     -     -     - 
Loose  Gun  Powder     - 
Spirits  of  Turpentine 
x\lcohol  or  Pure  Spirit 
Elm  and  Ash  -     -     - 
Oil  of  Turpentine  -     - 
Dry  Crab  Tree      -     - 

^ther 

White  Pine       -     -     - 
Sassafras  Wood  -     - 
Cork       -     -     -     -     - 
Common  Air  -     -     - 

Inflammable  Air    -     - 


2000 

2000 

1900 

1875 

1825 

1800 

1745 

1714 

1117 

1054 

1030 

1030 

1030 

lOOO 

993 

978 

925 

925 

913 

872 

864 

850 

800 

772 

765 

732 

569 

482 

240 


2-000 
2-000 
1-900 
1-875 
1-825 
1-800 
1-745 
1-714 
1-117 
1-054 
1-030 
1-030 
1-030 
I -000 
0-993 
0-978 
0n)25 
0-925 
0-913 
0-872 
0-864 
0-850 
0-800 
0-772 
0-765 
0-732 
0-569 
0-482 
0-240 
0-00125 

0-00012 


The  use  of  the  Table  of  Specifick  Gravities  will  best  appear  by 
several  Examples. 

How  to  discover  the  quantity  of  adulteration  in  metals. 

Suppose  a  body  be  compounded  of  gohl  and  silver,  and  it  be 
reqriied  to  find  the  quantity  of  each  nietaj  in  the  compound. 

First,  find  the  Specifick  gravity  of  the  compound,  by  weighing-  it 
in  air  and  in  water,  and  divivling  its  aerial  weight  by  what  it  loses 
thereof  in  water,  and  the  quotient  will  shew  its  specifick  gravity, 
or  how  many  times  heavier  it  is  than  its  bulk  of  water.  Then,  sub- 
tract the  specifick  gravity  of  silver  (found  in  the  Table)  from  that 
of  the  compound,  and  the  specifick  gravity  of  the  compound  from 
that  of  the  gold  :  the  first  remainder  will  j.hew  the  b-.jlk  of  gold, 
and  the  latter,  the  bulk  of  silver  in  the  whole  compound;  and  if 
these  remainders  be  multiplied  by  the  respective  specifick  gravi- 
ties, the  products  will  shew  the  proportional  weights  of  each  me*.« 
al  in  the  body. 


MISCELLANEOUS  QUESTIONS.  389 

Suppose  the  specifick  gravity  of  the  confipounded  body  be  14  ; 
that  of  standard  silver  (by  the  Table)  is  10  535,  and  that  of  standard 
gold  18-888  ;  therefore,  10-635  from  14,  remains  3465,  the  pro- 
portional bulk  of  the  gold  in  the  compound  ;  and  14  from  18-888, 
remains  4-888,  the  proportional  iw/A;  of  silver  in  the  compound: 
then,  18-888,  the  specifick  gravity  of  gold,  multiplied  by  the  first 
remainder  3  465,  produces  65-447  for  the  proportional  weight  of 
gold  ;  and  10  535,  the  specifick  gravity  of  silver,  multiplied  by  the 
last  remainder,  produces  51-495  for  the  proportional  weight  of  sil- 
ver in  the  vv^hole  body  :  So  that  for  every  65  447  ounces  or  pounds 
of  gold,  there  are  51-495  ounces  or  pounds  of  silver  in  the  body. 

Hence  it  is  easy  to  know  whether  any  suspected  metal  be  genu- 
ine, or  alloyed  or  counterfeit,  by  finding  how  much  heavier  it  is  than 
its  bulkof  water,  and  comparing  the  same  with  the  Table  ;  if  they 
agree,  the  metal  is  good ;  if  they  differ,  it  is  alloyed  or  counterfeited. 

How  to  try  Sjiirilous  Liquors. 

A  cubick  inch  of  good  brandy,  rum,  or  other  proofspirits,  weighs 
234  grains  ;  therefore  if  a  true  inch  cube  of  any  metal  weighs  234 
grains  less  in  spirits  than  in  air,  it  shews  the  spirits  are  proof:  If 
it  lose  less  of  its  aerial  wei<i;ht  in  spirits,  they  are  above  proof;  if  it 
lose  more,  they  are  under  proof;  for,  the  better  the  spirits  are, 
the  lighter  they  are,  and  the  worse,  the  heavier. 

Or,  let  any  solid,  of  sufficient  specifick  gravity,  be  weighed  first 
in  air,  then  in  water,  and  then  in  another  liquid  ;  from  its  weight 
in  the  air  take  its  weight  in  water,  and  the  remainder  is  the 
weight  of  its  bulk  of  water.  From  its  weight  in  air  take  its 
weight  in  the  other  liquid,  and  the  remainder  is  the  weight  of  the 
same  quantity  of  that  liquid.  Divide  the  weight  of  this  quantity  of 
liquid  by  the  weight  of  the  same  quantity  of  water,  and  the  quotient 
will  be  the  specifick  gravity  of  the  liquid. 

All  bodies  expand  with  heat  and  contract  with  cold  ;  but  some 
more,  and  some  less  than  others  :  therefore  the  specifick  gravities 
of  bodies  are  not  precisely  the  same  in  summer  as  in  winter. 

The  four  following  Problems,  relating  to    spiriious    liquors,  are 
wrought  by  Alligation, 

108.  What  proportion  of  rectified  spirits  of  wine  must  be  mixed 
with  water,  to  make  proof  spirit,  the  specifick  gravity  of  the  recti- 
fied spirits  being  850,  that  of  proof  spirit  925,  and  of  water  1000? 

OOP-  i  1000\75  )  ^  , 

>    850/75  V        equal  measures. 

109.  What  proportional  weight  of  rectified  spirits  of  wine  and 
water  must  be  mixed,  to  make  proof  spirit,  the  specifick  gravities 
as  before  ?  1000     20 

Ans. = — ,  or  as  20  to  17. 

850     17 

110.  What  is  the  specifick  gravity  of  best  French  brandy,  con- 
gisting  of  5  parts,  measure,  of  rectified  spirits  of  wine,  and  3  parts 
watpr  ? 


390  MISCELLANEOUS  QUESTIONS. 

850x5=4250 
1000x3=3000 

64-3=    8  )  7250 

906'25=specirick  gravity. 

111.  A  retailer  has  30  gallons  of  rum,  whose  speciiick  gravity 
is  900  t  How  much  water  must  he  add  to  reduce  it  to  standard 
proof? 

fjp-  (  1000 "NSB  }  '  g.  rum.  g.  wat.  g.  rum.  g.  wat. 

"^^,   900/75^  As  75     :     25     ::     30     ;     10  to  be  added. 

112.  The  cubick  inch  of  common  glass  weighs  about  l-36oz. 
Troy  :  ditto  of  salt  water  •5427oz.  ditto  of  brandy  •48927oz.  Sup- 
pose then,  a  seaman  has  a  gallon  of  brandy  in  a  bottle,  which 
weighs  4it{>  Troy,  out  of  water,  and  to  conceal  it,  throws  it  over- 
board into  salt  water :  Pray,  will  it  sink  or  swim,  and  by  how  much 
is  it  heavier  or  lighter  than  the  same  bulk  of  salt  water? 

54 
4i}fe=54oz.— weight  of  bottle  — -  =39-7059  cub.  in.  in  the  bottle. 

1  36 
Add  231"  =do.  in  the  brandy. 

270.7059=ditto  in  both. 
Then,  270  7059X'5427=146  912oz.=weight  of  salt  water  occu- 
pied by  the  bottle  and  brandy.  And  -48927  (=weight  of  a  cu- 
bick inch  of  brandy)  x231=l  1302+oz  and  11302+54=16702oz. 
=weight  of  the  bottle  and  brandy.  From  this  take  the  weight  of 
the  salt  water,  viz.  146192oz.  Ans.  Supposing  the  bottle  full,  it  is 
20II0Z.  heavier  than  the  same  bulk  of  salt  water,  and  therefore 
will  sink. 

Given  the  weight  to  be  raised  by  a  balloon^  to  find  its  diameter. 

Rule. 

1.  As  the  sprcific  difference  between  common  and  inflammable 
air,  is  to  one  oubick  foot :  so  is  any  weight  to  be  raised,  to  the  cu- 
bick feet  contained  in  the  balloon. 

2.  Divide  the  cubick  feet  by  '5236,  and  the  cube  root  of  the 
quotient  will  be  the  diameter  required,  to  balance  it  with  common 
air;  but,  to  raise  it,  the  diameter  must  be  somewhat  greater,  or 
the  weight  somewhat  less. 

113.  I  would  construct  a  spherical  balloon,  of  sufficient  capacity 
io  ascend  with  4  persons,  weighing,  one  with  another,  160ife,  and 
the  balloon  and  a  bag  of  sand  weighing  601fe  :  Required  the  diame- 
ter of  the  balloon  ? 

By  the  Table  of  Specifick  Gfravities,  page  388,  I  find  a  cubick 
foot  of  common  air  weighs  1  25  ounces  Avoirdupois,  and  a  cubick 
foot  of  inflammable  air  '12  of  an  ounce  Avoirdupois  ;  therefore. 


MISCELLANEOUS  QUESTIONS.  391 

Ife  ife       ft        oz. 

1  25— IS^MSoz.  difference.     And   160x4+60=700=11200. 
oz.  cub.  foot.  oz.      cub.  feet  3  9911-5044 

As  M3  :  1  ;:  11200  :  9911-5044.  And  V —=26. 65 feet, 

•5236  [Ans. 

Given  the  diameter  of  a  balloon^  to  find  what  weight  it  is  capable  of 

raising. 

Rule. 

1.  Multiply  the  cube  of  the  diameter  by  -5236,  and  the  product 
will  be  the  content  in  cubic  feet. 

2.  As  one  cubick  foot  is  to  the  specifick  difference  between  com- 
mon and  inflammable  air;  so  is  the  content  of  the  balloon  to  the 
weight  it  will  raise. 

114.  The  diameter  of  a  balloon  is  26  65  feet:  What  weight  is 
it  capable  of  raising  ? 

26-65x26-65X26-65X-5236i=9911-4-f  cubick  Ceet.     And 
cub.  foot.  oz.  cub.  feet.  oz. 

As   1     :      1-13  ::  9911  4-|-   :   1 1199882=7001fe  nearly. 
If  the  magnitude  of  any  body  be  multiplied  by  its  specifick  gra- 
vity, the  product  will  be  its  abi^olute  weight. 

115.  What  weight  of  lead  will  cover  a  house,  the  area  of  whose 
roof  is  6000  feet,  and  the  thickness  of  the  lead  yi^  of  a  foot  ? 
6000Xj^jf=50  cub.  feet,  and  its  specifick  gravity    11325X60= 

tons.  cwt.  qrs.   lbs.     oz. 
566250  ounces=15     15     3     26     10  Ans. 

To  find  the  magnitude  of  any  thing  Tvhen  the  weight  is  known. 

Divide  the  weight  by  the  specifick  gravity  in  the  Table,  and  the 
quotient  will  be  the  magnitude  sought. 

116.  What  is  the  magnitude  of  several  fragments  of  clear  glass, 
whose  weight  is  13  ounces  ? 

13-r-2600=-005  of  a  cubick  foot,  and  -005X1728=8.640  cubick 
!  inches,  Ans. 

Hating  the  magnitude  and  weight  of  any  body  given^  to  find  its  spe- 
cifick gravity. 

Divide  the  weight  by  the  magnitude,  and  the  quotient  will  be 
the  specifick  gravity. 

117.  Suppose  a  piece  of  marble  contains  8  cubick  feet,  and  weighs 
1353ilfe  or  21656  ounces  :   What  is  the  specifick  gravity  ? 

i       21656~-8=2707  the  specifick  gravity  required,  as  by  the  Table, 

^,  To  find  the  quantity  of  pressure  against  the  sluice  or  bank,  which  pens 

^'  water. 

Multiply  the  area  of  the  sluice,  under  water,  by  the  depth  of  the 
:entre  of  gravity,  (which  is  equal  to  half  the  depth  of  the  water)  iu 
feet;  and  that  product  again  by  H.^}  ftho  number  of  pounds  Avoir- 


392  MISCELLANEOUS  qUESTlONS. 

dupois  in  a  cubick  foot  of  fresh  water)  or  by  6 4-4lfe  (the'Avoiruu- 
pois  weight  of  a  cuhlck  foot  of  salt  water)  and  the  pr'oduct  will  be 
th*e  number  of  pounds  required. 

118.  Suppose  the  length  of  a  sluice  or  floom  be  30  feet,  the 
width  at  bottom  4  feet,  and  the  depth  of  the  water  4  feet;  what  is 
the  pressure  against  the  side  of  the  sluice  ? 

30X4=120  feet  the  area  of  the  bottom,  and  120X2  (the  depth 
of  the  centre  of  gravit}')  gives  240  cubick  feet,  and  240X62-5= 
15000ife=6T.  13cwt.  3qrs.  20Jfe  Ans. 

The  perpendicular  pressure  ofjluids  on  the  bottoms  of  vessels  is  estimat- 
ed by  the  area  of  the  bottom  multiplied  by  the  altitude  of  the  fluid. 

119.  Suppose  a  vessel  3  feet  wide,  6  feet  long,  and  4  feet  high, 
what  is  the  pressure  on  the  bottom,  it  being  filled  with  water  to  the 
brim? 

3X3=15  square  feet,  the  area  of  the  bottom,  and  16X4=60 
cubick  feet,  and  60X62-5=3750Ifei=t33  cwt.  1  qr.  261b. 

THE  USE  OF  THE  BAROMETER. 

The  Barometer  is  so  formed,  that  a  column  of  quicksilver  is  sup- 
ported within  it  to  such  a  height  as  to  counterbalance  the  weight  of 
a  column  of  air,  of  an  equal  diameter,  extending  from  the  barome- 
ter to  the  top  of  the  atmosphere. 

120.  At  the  surface  of  the  earth,  the  height  of  this  column  of 
quicksilver  is,  at  an  average,  almost  30  inches  ;  when  the  barom- 
eter is  at  that  height;  what  is  the  pressure  of  atmosphere  on  a 
square  foot,  and  on  the  surface  of  a  man's  body,  estimated  at  14 
square  feet? 

As  the  cubick  foot  of  quicksilver  is  13600  ounces.  Avoirdupois, 
and  as  the  height  in  the  barometer,  is  2'5  feet,  therefore  13600X25 
=34000  ounces,  =2125  pounds  on  a  square  foot ;  and  2125X14=^ 
29750  pounds  on  a  man's  body. 

121.  If  the  mercury  in  a  barometer,  at  the  bottom  of  a  tower,  be 
observed  to  stand  at  30  inches,  and,  on  being  carried  to  the  lop  of 
it)  be  observed  at  29*9  inches  :   What  is  the  height  of  the  tower  ? 

Divide  13600,  the  specifick  gravity  of  quicksilver,  by  1*25,  the 
specitick  gravity  of  air,  and  the  quotient  will  be  ti)e  height  of  the 
tower,  in  tenths  of  an  inch. 

13600  10880 

« =10880  tenths,  and =1088  inch.=90|  feet  Ans, 

1-25  10 

The  number  of  feet,  in  height,  of  the  atmosphere,  corresponding 
with  yV  of  an  inch  on  the  barometer  is  variable,  depending  oo  the 
temperature  and  density  of  the  atmosphere. 

The  variation,  depending  on  the  temperature,  is  shewn  in  the  fol- 
lowing Table,  calculated  for  every  5  degrees,  from  32  to  80,  Fahren- 
heit's Thermometer,  from  whence  it  may  be  easily  calculated,  for 
the  intermediate   degree*)  by  allowing  f  „^^  of  a  foot/or  each  degree. 


MISCELLANEOUS  QUESTIONS. 


393 


TABLE. 

Thermo.  Feet. 


32" 

86- 8G 

35 

87^49 

40 

88-54 

45 

89-60 

50 

90-66 

65 

91-72 

60 

92-771 

65 

93-82 

70 

94-88 

76  , 

95  93 

80 

96-99 

The  altitude, thus  found,  will  be  to  the  altitude  cor- 
rected for  the  density  of  the  air,  inversely,  as  the 
mean  height  of  the  barometer,  at  the  two  stations,  is 
to  30  inches  ;  therefore, 

Rule. — Multiply  the  mean  height  corresponding 
to  the  mean  temperature  ef  the  two  barometers 
(found  in  the  Table)  by  the  tenths  of  an  inch  in  the 
difference  of  the  two  barometers,  and  this  product  by 
30  ;  divide  this  last  product  by  the  mean  height  of 
the  two  barometers,  and  the  quotient  will  be  the  an- 
swer, or  height  required,  with  the  errour  of  a  few 
feet  only,  if  the  height  be  less  than  a  mile.* 


122.  At  the  first  station,  suppose  the  barometer  to  stand  at  29, 
and  the  thermometer  at  60  ;  at  the  second  station,  the  barometer 
at  28,  and  the  thermometer  at  40  :     What  is  the  height  of  the  2d 
station  or  the  distance  between  the  two  places  of  observation  ? 
Barometer. 
First  station      =29 
Second  station  =28 


Add 


i  sum==28'5=mean  height  of  the  two  barometers. 

29 
28 

Difference=  1  =  10  tenths  of  an  inch. 
Thermometer. 
First  station      =60 
Second  station  =40 

1)100 

60=mean  height  of  the  two  thermometers,  against 
which,  in  the  Table  you  will  find  90-66,  the  mean  temperature  of 
the  two  barometers.  Now,  according  to  the  rule  90  66x  10x30~ 
28'5==954-3feet,  the  Answer,  nearly. 

*  Let^  =  mean  height  of  the  barometer  at  its  two  stations,  (or  of  two  barom- 
eters, one  at  each  station)  in  inches ;  d  =  difference  of  the  two  barometers  in 
tenths  of  an  inch ;  and  n  =z  number  from  the  Table  answering  to  the  mean  tem- 

30dn 

perature  of  the  two  thermometers  accompanying.the  barometer,  then =  the 

h. 
altitude  required  nearly. 


B  3 


394  OF  THE  VALUE  OF  COINS. 

The  Act  of  Congress  of  April  29,  1816,  regulating  the  cnrrency 
within  the  United  States  of  the  gold  coins  of  Great  Britain,  France, 
^c.  enacted, 

That,  of  the  gold  coins  of  Great  Britain  and  Portugal, 
27  grs.=:100  cents,  or  1  pwt.—80f  cents  ; 
Of  France,  27igrs.=  do.  do.  =87^     do. 

Spain,     28igrs.=  do.  do.  =84       do. 

Crowns  of  France,  weighing  449grs.  =  irOcenls,  or  loz.  =  117c. 

Five  franc  pieces,  weighing  386grs. =93-3  cts.  or  loz.=ir6cts. 

The  Spanish  dollar,  weighing  not  less  than  416grs.==100  cents. 

The  Federal  dollar  is  to  be  of  the  weight  and  purity  of  the 
Spanish  dollar;  but,  it  is  to  weigh  416grs.  and  to  contain  371^grs. 
of  pure  silver. 

The  pound  Sterling  of  England  is  ^4*44  in  the  United  States. 
The  dollar  is  reckoned,  therefore,  at  four  shillings  and  six  pence 
sterling;  but  in  1820,  four  shillings  and  six  pence  of  English  silver 
coin  contained  only  363^grs.  of  pure  silver,  being  8grs.  less  than  ia 
contained  in  the  Federal  dollar. 

One  pound  Troy  weight  of  Standard  Gold  in  England,  contains 
6280grs.  of  pure  gold,  and  is  coined  into  £4Q  14s.  6d.  or  11214 
pence.     Hence  a  pound  sterling  contains  1 13  0016grs.  of  pure  gold. 

'J'he  Eagle  contains  247-5grs.  of  pure  gold,  and  hence  the  pound 
sterling  is  worth  in  gold  ^4  56-572,  and  hence  the  dollar  is  worth 
in  English  gold  4s.  4-5656d. 

.  One  pound  Troy  of  Standard  silver  in  England  contains  6328grs. 
of  pure  silver,  and  is  coined  into  66  shillings  or  792  pence.  As  the 
dollar  contains  37Jigrs.  of  pure  silver,  the  dollar  is  worth  in  Eng- 
lish silver  4s.  71858d. 

In  a  pound  sterling  there  is  1614  545grs.  of  pure  silver,  which 
is  equal  in  silver  to  §4-34  8943. 

Taking  the  mean  of  the  values  of  the  dollar  and  the  pound  ster- 
ling in  gold  and  silver, 

The  value  of  the  dollar  is  4s   5-8757d.  sterling. 

And  the  value  of  the  pound  sterling  is  g4-45-7331. 

This  mean  value  of  the  dollar  and  the  pound  sterling  is  very  near 
the  values  at  which  they  are  commonly  estimated. 

[See  ^'  Report  of  the  Secretary  of  Slate  upon  Weights  and  Meas- 
ures," appendix  C.  to  Congress,  Feb.  1821.] 

The  standard  price  of  gold  in  England,  is  £3  17s.  10|d.  an  oz. 
and  of  silver  5s.  2d.  an  oz.  The  standard  weight  of  the  English 
jguinea  is  5pwt.  9|grs. ;  but  it  usually  weighs  5pwt.  8grs. 

The  standard  coin  of  France  is  to  contain  one  tenth  of  alloy,  and 
the  standard  value  of  gold  to  silver  is  15  to  1. 


TABLES. 


395 


The  following  Tables  are  calculated  according  to  the  value  of 
Foreign  Gold  established  by  act  of  Congress,  April  29,  1816. 


TABLE  I. 


TABLE  II. 


V^alue  of 

LCngl] 

sh  and   1- 

ortuguese 

Gold,  in 

dollars,  cents, 

and  mills, 

throug:hout  the  United  States.        | 

Gr. 

Ctg, 

m. 

Pwts. 

$    cts. 

1 

3 

7 

1 

0  881 

2 

7 

4 

2 

1   77f 

3 

11 

1 

3 

2  66| 

4 

14 

8 

4 

3     55 

5 

181 

0 

5 

4     44 

6 

22 

2 

6 

5  331 

7 

25 

9 

7 

6  32 

8 

29 

6 

8 

7   11 

9 

331. 

0 

9 

8     0 

10 

37' 

0 

10 

8  89 

11 

40 

7 

11 

9  77| 

12 

44 

4 

12 

10  G6 

13 

48 

1 

13 

11   551 

14 

51 

8 

14 

12  44 

15 

55 

5 

15 

13  331 

16 

591 

0 

16 

14  22 

17 

63^ 

0 

17 

15   11 

18 

GGs. 

0 

18 

16     0 

19 

70' 

4 

19 

16  89 

20 

74 

0 

loz. 

17  77| 

21 

771 

0 

J^ote. 

88  8.  cents, 

22 

8lt 

0 

the  value 

of  1  pen- 

23 

85^ 

2 

ny-weigh 
and  Port 

it  of  Eng. 
ugs.  Gold. 

Value  of  French  Gold,  in  dolls,  cents, 

and  mills,  in  the  United  States. 

Gr. 

Cts.  m. 

Pwt. 

$    Cts. 

1 

3  63. 

1 

0  87i 

2 

7  3 

2 

1   741 

3 

10  9 

3 

2  613 

4 

14  5/- 

4 

3  49 

5 

18  2    ' 

5 

4  361 

6 

21   8 

6 

5  231 

7 

25  4-6- 

7 

6   10| 

8 

29  r^ 

8 

6  98 

9 

32  7_3_ 

9 

7  851 

10 

36  3_7. 

10 

8  721 

11 

40  0 

11 

9  59| 

12 

43  6-^j 

12 

10  47 

13 

47  3 

13 

11   341 

14 

50  9 

14 

12  211 

15 

54  51 

15 

13  083 

16 

58  2 

16 

13  96 

17 

61   8 

17 

14  831 

18 

65  4-6^ 

18 

15  701 

19 

69   l'' 

19 

16  57| 

20 

72  7 

loz. 

17  45 

OJ 

76  3  ^ 

22 

80  Q 

Mte. 

87i  cents 

is   the    V 

alue   of  1 

23 

83  6A 

pwt.  of  ] 

^re.  Gold. 

K)  K)  ^« 

.^   ^_   ^^   ^_.                                                                  /^ 

so 

Ococo->3050T*^03^s»-ooc^<^0:OT*-oo^o^-7f 

C0<i<{^-|C5a)CnOxCn 

Ml-,  toi^  wl-  Ml-  laj- 

CD 

1— t 

cn60cocn  —  corfi.>--a*»0<i00co' 
a>i-*Oi»-*cr'OCnOCr<OOiOCnB 

4>|M  M|M  00|Ol            I0|-  Mj—  *-|—                         r 

o 

CD    *-» 

^^ 

H-  CO  CO  <{  oi  en 
o 

N 

*^ 

03tsi»— ococ5<la)0^4;^oo^^>— ^ 

12  60 

13  44 

14  28 

15  12 

15  96 

16  80 

^1 

Ci 

OOCOCO<IOiCnCn4^00tO^  O"^ 

co0^o►;^OT•>3coo^oo^cna)Coo 
toco4^oa>^^co►^oo5J^co4-J* 

396 


TABLES  OF  EXCHANGE. 


N.  Hamp- 

New    Jer- 

shire, 

New  York 

sey,    Penn- 

South Caro- 

Canada 

Federal 

Massachu- 

and 

sylvania, 

lina  and 

and  Nova-  French. 

Money. 

setts,R.Isl- 
and,  Con. 

North  Ca- 
rolina. 

Delaware, 
and  Mary- 

Georgia. 

scotia. 

feVirginia. 

land. 

*;    c. 

X,    s.    d. 

Jj.     s.  d. 

£     s.     d. 

£        s.     d. 

£      £.     d. 

^     Sou. 

00  1 

¥s 

II 

T^o 

if 

1 

iii 

00  2 

m 

m 

lA 

h% 

n 

0  0  3 

2oV 

22.2. 
^25 

2t\ 

n^ 

H 

■^     3^^. 

00  4 

2|2 

m 

3t\ 

2/. 

22 

4^ 

0-0  6 

3| 

4  f 

4fV 

2| 

3 

5  I 

0-0  6 

4^ 

m 

^j\ 

3/j 

^ 

6^ 

00  7 

^oV 

m 

6f^. 

m 

H 

^.^ 

0-0  8 

^M 

m 

7^ 

m 

^ 

s  ! 

0-0  r 

m 

m 

^tV 

Ni 

5! 

9^\ 

0  1  0 

^  i 

^1 

9 

^  f 

6 

1^  i 

• ,  1  ■ 

1     2     2 

1  n 

2  41 

1   6 

1    -4  1 

1      10     2 

1   0 

1      1 

0-3  0 

1   9| 

2  3 

1   6 

1    H   i 

0-4  0 

2  4   4 

3  2l 

3  0 

2  0 

2     2 

0-5  C 

3  0  ' 

4  O' 

3  9 

2     4   ' 

2  G 

2    12  1 

Ob  0 

3  7  1 

4  9| 

4  6 

2     9^ 

3  0 

3     3 

07  0 

4  2  f 

6  7;l 

5  3 

3     3  i 

3  6 

3  13  X 

4  4 

0  8  C 

4  9  1 

6  4f 

6  0 

3  8  i 

4  2  1 
4     8 

4  0 

\ 

0-9  0 

6  4  A 

7  2| 

6  9 

4  6 

4   14  1 

\ 

10  0 

6  0 

8  0 

7  6 

5  0 

5     5 

2-0  0 
3-0  0 
4-0  0 

12  0 

16  0 

15  0 

9     4 

10  0 

10  10 

18  0 

1     4  0 

12  6 

14    0 

15  0 

15  15 

14  0        1  12  0 

1   10  0 

18     8 

1     0  0 

21     0 

rrO  C 

1   10  0 

2    0  0 

1   17  6 

1     3     4 
1     8     0 

1     5  0 

26     5 

6-0  0 

1  16  0 

2     8  0 

2     5  0 

1   10  0 

31  10 

7-0  C 

2     2  0 

2  16  0 

2  12  6 

1    12     8 

1  15  0 

36  15 

8-0  0 

2    8  0 

3    4  0 

3    0  0 

1  17     4 

2    0  0 

42    0 

9-0  0 

2  14  0 

3  12  0 

3     7  6 

2     2    0 

2     5.0 

47     5 

10-0  0 

3    0  0 

4    0  0 

3  15  0 

2     6     8 

.    2  10  0 

52  10 

20-0  0 

6     0  0 

8     0  0 

7   10  0 

4  13    4 

5     0  0 

105     0 

30-0  0 

9     0.0 

12     0  0 

11     5  0 

7     0    0 

7  10  0 

157  10 

40-0  0 

12    0  0 

16     0  0 

15     0  0 

9     6     8 

10    0  0 

210     0 

50-0  0 
GOO  0 

15     0  0 
18     0  0 

20     0  0 

18  15  0 

11   13    4 
14     0     0 

12  10  0 

262  10 

24     0  0 

22  10  0 

15    0  0 

315     0 

70-0  ( 

21     0  0 

28     0  0 

2G     5  0 

16     6     8 

17  10  0 

367  10 

r    {<o-o  0 

24     0  0 

32     0  0 

30     0  0 

18  U    4 

20    0  0 

420    0 

90-0  (J 

27     0  0 

36     0  0 

33  15  0 

21     0    0 

22  10  0 

472  10 

1    100-0  ( 

30     0  0 

40     0  0 

37   10  0 

23     6     8 

25     0  0 

525     0 

'200-0  ( 

60     0  0 

80     0  0 

75     0  0 

46  13    4 

50     0  0 

1050    0 

300-0  C 

90     0  0 

120     0  0 

112  10  0 

70    0    0 

75     0  0 

1575     0 

!  400-0  c 

120    0  0 

160     0  0 

150     0  0 

93     6     8 

100    0  0 

2100    0 

500-0  t 

)  150    0  0 

200     0  0 
240     0  0 

187  10  0 

116  1.^    4 

125     0  0 

[2625     0 

boo-0  ( 

)  180     0  0 

225     0  0 

140     0     0 

150    0  0 

3150     0 

700-0  ( 

)  210     0  0 

280    0  0 

262  10  0 

163     6     8 

175     0  () 

3675    0 

000-0  ( 

)  240     0  0 

320    0  0 

300    0  0 

186  13    4. 

200    0  0 

42G0    0 

900-0  ( 

)  270     0  0 

360    0  0 

337  10  0 

210     0    0 

225     0  0 

4725     0  I 

1000-0  ( 

)  300    0  0 

MOO    0  0 

375     0  0 

233     6     8 

250    0  0 

5250     0 

TABLES  OF  EXCHANGE. 


^97 


' 

N.  Ham. 

N.  Jersey, 

~~^ 

1 

•——— 

r 

Mass. 

Federal 

New  York 

Pennsyl- 

S.Carolina i 

English 

French 

R.  Island, 

Coin. 

and  North 

vania,  De- 

and 

Money. 

Money. 

Con.  and 

Carolina. 

laware,  & 

Georgia.     1 

Virg;inia. 

Maryland. 

d. 

£      s.  d. 

$     Cts. 

£         3.      d. 

Jj     s.     d. 

£      s. 

£ 

s.  d. 

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12    0  0 

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157  15 

1    10  0  0 

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13    6  8 

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7  15 

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7 

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175    0 

20  0  0 

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26  13  4 

25    0  0 

15  11 

H 

15 

00 

350    0 

30  0  0 

100  00 

40    0  0 

37  10  0 

23    6 

8 

22 

100 

525    0 

;  40  0  0 

133-331 

53    6  8 

50    0  0 

31     2 

Q6 

30 

00 

700    0 

50  0  0 

166661 

GG  13  4 

62  10  0 

38  17 

^ 

37 

10  0 

875    0 

CO  0  0 

200  00 

80    0  0 

75    0  0 

46  13 

4 

45 

0  0 

1050    0 

j   70  0  0 

233-331 

93    6  8 

87  10  0 

54     8 

10^ 

52 

10  0 

1225    0 

80  0  0 

266  66| 

106  13  4 

100    00 

62     4 

H 

60 

00 

1400    0     ] 

90  0  0 

300-00"' 

120    00 

112  100 

70    0 

0 

67  10  0 

1575    0 

!  100  0  0 

333-33J 

133    6  8 

125    0  0 

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0  0 

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666  66-;;- 

266  13  4   1250    0  0 

155  11 

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00 

3500    0 

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100000 

400    0  0    375    0  0 

233    6 

8 

1225 

00 

5250    0 

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1333-33 

533     6  8    500    0  0 

311     2 

25 

300 

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7000    0 

500  0  C 

1666-66 

666  13  4 

1625    0  0 

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375 

0  0 

8760    0 

598 


TABLES. 


TABLE 

©F   THE   VALUE  OP   SEVERAL   PIECES    OF    COIN,  IN   THE   FEDERAL  COIN,  AND 
THE   SEVERAL    CURRENCIES    OF   THE    UNITED    STATES. 


N.  Hamp. 

N.  Jersey, 

Federal 

Mass. 

New  York 

Pennsylva- 

S. Carolina 

Coin. 

R.  Island, 

and  North 

nia,    Dela- 

and 

Con.  and 

Carolina. 

ware  and 

Georgia. 

Virginia. 

Maryland. 
£    s.     d. 

Cents. 

£     s.  d. 

£    s. 

d. 

J^      s.    d. 

J^ofa  Dollar 

0-06   1 

n 

6 

5f 

^ 

iofa  Pistareen 

0-10 

Vir.    8 

9| 

9 

H 

iofa  Dollar 

0-11   1 

8 

10| 

10 

6f 

i  of  ditto 

012  1 

9 

1 

0 

iH 

7 

A  Pistareen 

020 

1   2| 

1 

6 

1     6 

Hi 

An  Eng.  Shilling 

022  1 

1   4 

1 

7f 

1     8 

1     Of 

1  of  a  Dollar 

0-25 

1   6 

2 

0 

1    101 

1      2 

Half  ditto 

0-50 

3  0 

4 

0 

3     9' 

2     4 

A  Dollar 

100 

6  0 

8 

0 

7     6 

4     8 

French  Crown 

110 

6  7 

8 

9-} 

8     3 

5     U 

pwt.  gr. 

! 

fr.  Guineas     6 

4-64  I 

I     7  31 

1    16 

n 

1    14     0 

1     1     21 

En.Guinea  6     6 

4-66  f 

1     8  0 

1    17 

4 

1    15     0 

1     1     31 

1  Johann.  9     0 

8-00 

2     8  0 

3     4 

0 

3     0     0 

1    17     4 

Pistole        4     5 

3-53  111      1   2i 

1     8 

3 

1     6     6 

16     5 

Moidore     6   18 

600 

1    16  0 

2     8 

0 

2     5     0 

1      8     0 

Donbloonl7     0 

14  66  1 

4     8  0 

5   16 

0 

5   12     0 

3   10     0 

Blanks 

24  ==   1  Perrot. 
480  =  20  ==    1  Mite. 
9600  —  400=  20  =   1  Gram. 


Table  of  Refiner's  Weights. 

J>foie.  What  they  denominate  a 
carat  is  the  ^'^  of  a  !fe  an  oz.  or 
any  other  weight. 


Dutch  Weights  for  Gold   and  Silver. 

Note,  32aces=l  engel,  20  engels=l  ounce,  8  ounces=lmark, 
for  jrross  gold.  Also,  24  parts=l  grain,  12  grain8=l  carat,  24 
farats=l  mark,  for  fine  gold. 

The  mark  weights  are  1  per  cent,  lighter  than  our  Troy  weight. 


TABLES. 


A  TABLE  OF  COMMISSION  OR  BROKERAGE. 


399 


Goods  or 
stock  sold 


Shut.   1 

2 

.  .  3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

\Pounds 1 


9 

10 

20 

30 

40 

60 

60 

70 

80 

90 

100 

200 

300 

400 

500 

600 

700 

800 

900 

1000 


at  i  per 
cent. 


0 
0 
0 

0  0 
0 


0  0 

0  0 

0  0 

0  0 

0  0 

0  0 


s.  d. 

0  0, 

0  04 

0  04 

0  Oi 

0  Of 

0  oi 

0  oi 

0  ol 

0  oi 

0  0  oi 

0  oh 

0  01 

0  Of 


Of 
0     0| 

1 


0 

0 
0  0 
0  0  81 
0  0  91 
0     0  lOf 

1  0 

2  0 
3 
4 

0  5 
0  6 
0  7 
0  8 
0     9 

0  10     0 

1  0     0 

1  10     0 

2  0     0 

2  10     0 

3  0     0 

3  10     0 

4  0     0 

4  10     0 

5  0     0 


at  1  per  | 

c 
0 

ent. 

0  0 

0 

0  Oi 

0 

0  04 

0 

0  0^ 

0 

0  0^ 

0 

0  01 

0 

0  0| 

0 

0   1 

0 

0  1 

0 

0  n 

0 

0  u 

0 

0   M 

0 

0  u 

0 

0  11 

0 

0   11 

0 

0  2 

0 

0  2 

0 

0  2A 

0 

0  n 

0 

0  2h 

0 

0  5 

0 

0  7 

0 

0  9i 

0 

1   0 

0 

1   2i 

0 

1   41 

0 

1  7 

0 

1  9^ 

0 

2  0 

0 

4  0 

0 

6  0 

0 

8  0 

0 

10  0 

0 

12  0 

0 

14  0 

0 

16  0 

0 

18  0 

1 

0  0 

2 

0  0 

3 

0  0 

4 

0  0 

5 

0  0 

6 

0  0 

7 

0  0 

8 

0  0 

9 

0  0 

10 

0  0 

at  H  per 
cent. 


0   1, 


0  0  0 

0  0  04 

0  0  Oh 

0  0  Oi 

0  0  Oj 

0  0  1 

0  0  u 

0  0  u 

0  0  H 

0  0  1* 

0  0  II 

0  0  2 

0  0  24 

0  0  2i 

0  0  2^ 

0  0  2^ 

0  0  3 

0  0  3 

0  0  3i 

0  0  34 

0  0  7i 

0  0  lOf 


at  2  per 
cent. 


0    12      0 


0 


0    18      0 


1  1  0 

1  4  0 

1  7  0 

1  10  0 

3  0  0 

4  10  0 

6  0  0 

7  10  0 
9  0  0 

10  10  0 

12  0  0 

13  10  0 
15  0  0 


0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 

1 
1 
1 
1 
1 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 


0  0 
0  04 
0  0^ 
0  Of 
1 


0 

0  14 

0  u 

0  1| 
0  2 
0  24 


0  2* 
0  3 
0  3i 
0  3h 
0  31 
0  4 
0  4i 
0  4i 
0  4| 
0  9^ 


1 
1 
2 
2 
2 
3 
3 
4 
8  0 
12  0 
16  0 
0  0 
4  0 
8  0 
12  0 
16  0 
0  0 
0  0 
0  0 
0  0 
0  0 
0  0 
0  0 
0  0 
0  0 
0  0 


at^ 

li  per  1 

cent.       j 

0 

0  04 

0 

0  0^ 

0 

0  0| 

0 

0  1 

0 

0  n 

0 

0  u 

0 

0  2 

0 

0  24 

0 

0  2h 

0 

0  3 

0 

0  3^ 

0 

0  3i 

0 

0  31 

0 

0  4 

0 

0  4i 

0 

0  4i 

0 

0  5 

0 

0  54 

0 

0  5i 

0 

06 

0 

1  0 

5 

7 
10 
12 
15 
17 
20 
22 
25 


1  6 

2  0 

26 

30 

36 

40 

46 

50 

10  0 

15  0 

00 

5  0 

10  0 

15  0 

00 

5  0 

10  0 

00 

10  0 

00 

10  0 

00 

10  0 

00 

10  0 

00 

at  3  pe" 
cent. 


0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 

1 

1 
1 

2 
2 
2 
3 
6 
9 

12 

15 

18 

21 

24 

27 

30 


0  0 

0  Oh 
0  1 
0  l4 
0  II 
0  2 

0  24 

0  2| 
0  3 
0  3^ 
0  31 

0  44 

0  4i 
0  5 
0  54 

0  5I 

0  6 
0  64 
0  6^ 

0  7 

1  2-2 

1  9h 

2  41 

3  0 

3  7 

4  21 
4  91 
6  4i 
6  0*^ 

12  0 
18  0 

40 
10  0 
16  0 

2  0 

80 
14  0 

00 

00 

00 

00 

00 

00 

00 

00 

00 

00 


400 


TABLES. 


A  TABLE 


/  THE  RETURNS  OF  THE  NEAT  PROCEEDS  OF  AN  ACCOrNT  OF  SALES 
/FROM  A  FACTOR  TO  HIS  EMPLOYER,  RESERVING  HIS  COMMISSIONS  FOR 
/  REMITTANCE. 


Sum  to  be 

Sum  to  be 

Sum  tobel  Sum  to  be! 

NeatP 

re- 

remitted, 

remitted, 

Neat  pro 
ceeds. 

remitted, 

remitted, 

ceedi: 

reserving 

reserving 

"  reserving  2i|reservmg  5j 

2i  per  ct. 

5  per  ceni 

per  ct.  com 

-  per  ct.  com- 

commisn. 

commisn. 

mission. 

mission. 

£     s. 

el.  JC    s.    d. 

£   s.    d. 

£    s.d. 

£     s.     d. 

£      s.     d. 

3 

3 

21 

6  0  0 

5  17    Of 

5  14     5i- 

4 

4 

31 

7  0  0 

6  16    7 

6  13     4 

j 

5 

5 

4f 

8  0  0 

7  16     H 

7  12    4i 

6 

51 

51 

9  0  0 

8  15     7i 

8  11     5i 

7 

61 

6i 

10  0  0 

9  15     U 

9  10     51  3 

8 

71 

7i 

20  0  0 

19  10    3 

19    0  lU   1 

9 

81 

8i 

30  0  0 

29     5     4^ 

28  11     5i  \ 

10 

91 

9^ 

40  0  0 

39    0    5^ 

38  1  m  \ 

11 

101 

101 

50  0  0 

48  15     7i 

47  12    4i 

1 

0 

111 

lU 

60  0  0 

58  10     8| 

57     2  lOi 

2 

0 

1  lU 

1   101 

70  0  0 

68     5  10 

66  13    4 

3 

0 

2  iH 

2  m 

80  0  0 

78     0  Ul 

76     3     9| 

4 

0 

3  10| 

3"   91 

90  0  0 

87  16     1 

85  14    3i 

5 

0 

4  1O.5 

4     9^ 

100  0  0 

97  11     21 

95     4     9 

6 

0 

3  104 

5     8i 

200  0  0 

195     2    51 

190    9    6i 

7 

0 

6  10 

6     8 

300  0  0 

292  13     8 

285  14    3i 

8 

0 

7     91 

7    7i 

400  0  0 

390    4  10^ 

380  19    Oi 

9 

0 

8     9i 

8    6| 

500  0  0 

487  16     H 

476     3    94 

10 

0 

9     9 

9     6i 

600  0  0 

585     7     31 

571     8     6| 

1     0 

0 

19     6i 

19    0^ 

700  0  0 

682  18     4| 

666  13     4 

2    0 

0 

1  19    Oi 

1  18     H 

800  0  0 

780     9     9 

761  18     1 

^    0 

0 

2  18     6'^ 

2  17     1| 

900  0  0 

878    0  111 

857     2  10 

4    0 

0 

3  18    Oi 

3  16    2i 

1000  0  0 

975  12    2i 

952    7    7i 

5     0 

0 

4  17     6^ 

4  15    21 

Suppose  I  have  the  neat  proceeds,  or  balance  of  an  account  of 
sales  3251.  17s.  9d.  in  my  hands  and  would  make  remittance  to  my 
employer,  reserving  my  commission  at  2i  per  cent.  What  sum 
must  be  remitted,  so  that  my  employer's  account  may  be  closed  ? 


Against. 


£ 

s.   d. 

fSOO 

0  0~] 

20 

0  0 

5 

0  0 

10  0 

7  0 

9 

£ 

c. 

d. 

292 

13 

8 

19 

10 

3 

4 

17 

61 

9 

9 

6 

10 

8| 

S  stands. 


To  be  remitted  £317  18  9i  Answer. 


TABLES. 
A  TABLE, 


40i 


WiEWliVG    THE   arUMBER    OF   DAYS   FROM   ANT    DAT    IN   ANY   MONTH  TO  THE 
S13IE   DAY    ly    ANY    OTHER    MONTH    THROUGH    THE    YEAR. 


From  Jan  Feb  Mar  Ap  May  Juu  Jul  Aug  Sep  Oot  Nov  Dec| 

To  Jan. 

365 
"3! 
59 
"90 
120 
131 
181 
212 
243 
273 
304 
334 

334 
365 
T8 
'59 
Tl' 
120 
150 
181 
212 
242 
273 
303 

306 

337 

365 

3] 

61 

92 

122 

153 

184 

214 

245 

275 

306 

334 

365 

30 

61 

9\ 

^ 

153 

183 

214 

24^ 

245 
276 
304 
335 
365 
Tl 
6l 
'92 
123 
153 

214 

214 

24f: 

304 
334 
365 

30 

~6i 

— 
92 

122 

153 

183 

18. 
215 
243 
274 
.304 
335 
36^ 
31 
~62 
~92 
123 
153 

153 
184 
,212 
243 
273 
304 
334 
3^:5 
31 
61 
T2 
122 

122 
\53 

iFi 

212 

242 

273 

303 

334 

365 

30 

61 

91 

9,i 
l23 
151 
182 

2r2 

243 
273 
304 
335 
365 
31 
61 

6i 
92 
120 
151 
181 
2\2 
242 
273 
304 
334 
365 
30 

31 
62 
"90 
1211 
151 
182 
2r2 
243 
274 
304 
335 
365 

Feb. 

Mar, 

Apr. 
May. 

June. 

July. 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

The  Use  of  the  preceding  Table  of  number  of  days,  will  easily 
ippear  from  the  following  examples. 

Suppose  the  number  of  da^s  between  the  first,  or  10th,  or  30th, 

&c.  of  January,  and  the  1st,  or  10th  or  30th,  &c.  of  October,  were 

required  :     Look  in  the  column  under  January  for  October,  and 

I   against  that  month  you  will  find  2^73,  which  is  the  number  of  days 

^  between  the  said  times;  and  so  for  the  days  between  any  other 

two  months. 

If  the  given  days  be  different,  it  is  only  adding  or  subtracting  their 
inequality  to  or  from  the  tabular  number. 

How  many  days  from  the  6th  of  April  to  the  12th  of  January  ? 
From  the  6th  of  April  to  the  6th  of  January  is  275,  and  adding  the 
6  overplus  days,  it  makes  281  days.  And  from  the  6th  of  June  to 
the  1st  of  February  is  240  days. 

Note.  After  February  31  (in  leap  years)  increase  each  num- 
ber with  an  unit  or  1. 


rj    o, 


U)^  TABLES. 

A  Table  of  the  Measure  of  Length  of  the  principal  places  in  Europe, 
compared  txith  the  American  yard. 

100  Aunes  or  Ells  of  England,  -         ,         -         -         =  125 

100  of  HoUand  or  Amsterdam,  Haerlem,  Leyden^ 

the   Hague,  Rotterdam,  Nuremberg,  and  >      =     75 
other  cities  of  Holland,  7 

100  ~ of  Barbant,  or  Antwerp,  -         -         -  =-76 

100 of  France  and  Oznaburg,  -  -         -         =   1284- 

100  ofHamburg,  Frankfort,  Leipsick, Bern,  &  Basil,  =     62^^ 

100  ofBreslau, =60" 

100  ofDanlzick,  ,....=     66f 

100 of  Bergen  and  Drontheim,         ,         .         _         =     68i 

100  of  Sweden  and  Stockholm,         .         -         -  =     65| 

100  of  St.  Gall,  for  linens,       .         -         -         -         =     87| 

100  of  ditto  for  cloths,  -         -         -         .         =     67 

100  of  Geneva,  =124| 

100  .Canes  of  Marseilles  and  Montpelter,  -         -         =214^ 

100  ofThoulouse  and  High  Languedoc,  -         -         =  200" 

100  of  Genoa,  of  9  palms,       -.-.=:  245i 

100  of  Home,         - =  227i 

100  Varas  of  Spain,       -         -         -         -         .         -         =     93i 

100  of  Portugal,     -  -  ,  .         -  -  =123 

100  Cavidos  of  Portugal, =75 

100  Brasses  of  Venice,  ._«--=     73^ 

100  of  Bergamo,  -         -         -         -         -         =     7li 

100  of  Florence  and  Leghorn,  -         "         '         =64 

100  of  Milan,  ..-..-=     58^ 

The  use  of  the  follomng  Table ,  directiitg^  how  to  buy  and  sell  by  the 

hundred. 

If  you  buy  or  sell  any  thing  by  the  great  hundred  (llSlfe)  and 
desire  to  know,  .by  the  fe,  what  the  hundred  is  valued  at,  observe 
the  following  examples. 

1.  If  you  buy  sugar  at  GU.  per  ife,  look  for  6ld.  in  the  left  hand 
column  of  the  Table,  against  it  in  the  second  column,  you  will  find 
£3  3s.  which  is  the  value  of  Icwt.  at  that  rate. 

2.  If  Icwt.  (112Jfe)  cost  £9  4s.  4d.  to  know  how  much  it  is  pet* 
ife,  look  £9  4s.  4d.  in  the  tburth  column,  and  against  it  in  the  next^ 

-left  hand  column,  you  will  find  Is.  7ld.  which  is  the  price  per  S5. 

Again,  If  you  buy  one  hundred  weight  of  goods  for  91.  4s.  4d.| 
and  retail  it  at  Is.  9|d.  per  ife,  it  comes  at  that  rate  to  101.  3s. ;  thei^ 
take  91.  4s.  4d.  from  101.  .3s.  am!,  by  the  remainder,  you  will  findj 
that  you  have  gained  18s.  8d. 

And  in  this  manner  you  may,  with  ease,  cilculate  any  quantityl 
by  the  following  Table. 


TABLES. 


40? 


A  ^ABLE    DIRECTmc    HOW   TO   BUY   AND   SELI,   BY   THE   HUNDRED. 


d. 

£ 

s.  d. 

s.  d. 

i;  s.  d. 

s.   d. 

£    s.  d. 

i 

0 

2  4 

1   Oi 

6  14  4 

2  01 

116  4 

^ 

0 

4  8 

1   01 

5  )6   8 

2  01 

11   8  8 

f 

0 

7  0 

1   0| 

5  19  0 

2  01 

11  11  0 

1 

H 

0 

9  4 

1   1 

6   1  4 

2     1 
2   11 

11  13  4 

0 

11  8 

1   ^i 

6  3  ( 

11  15  8 

n 

0 

14  0 

1   1^ 

6  6  0 

2   ]i 

11  18  0 

H 

0 

IG  4 

1   If 

6  8  4 

2  H 

12  0  4 

2 

0 

18  8 

1  9 

6  10  8 

2  2 

12  2  8 

^i 

1  0 

1   iii 

G  13  0 

2  21 

12  5  0 

n 

3  4 

1   21 

6  15  4 

2  21 

12  7  4 

21 

5  8 

1   21 

6  17  8 

2  21 

12  9  8 

3 

8  0 

1  3 

7  0  0 

2  3 

12  12  0 

H 

10  4 

'   31 

7   2  4 

2  31 

12  14  4 

31 

12  8 

1   31 

7  4  8 

2  3 . 

12  16  8 

31 

^ 

15  0 

1   31 

7  7  0 

2  3| 

12  19  0 

4 

1 

17  4 

.1   4 

7  9  4 

2  4 

13  1  4 

44 

19  8 

1   44 

7  11  8 

2  44 

13  3  8 

4i 

2 

2  0 

1   4i 

7  14  0 

2  41 

13  6  0 

4f 

2 

4  4 

1  4^ 

7  16  4 

2  41 

13  8  4 

5 

2 

6  8 

1   5 

7  18  8 

2  6 

13  10  8 

■  ^} 

2 

9  0 

1  r.! 

8   1  0 

2  51 

13  13  0 

5h 

2 

n  4 

1  «^4 

8  3  4 

2  5l 

13  15  4 

5f 

2 

13  .8 

1   51 

8  5  8 

2  5| 

13  17  8 

6 

2 

16  0 

1  G 

8  8  0 

2  6 

14  0  0 

6i 

2 

18  4 

1   C)i 

8  10  4 

2   61 

14  2  4 

4 

3 

0  8 

1  61 

8  12  8 

2  61 

14  4  8 

6l 

3 

3  0 

.  1  6l 

8  15  0 

2  61 

14  7  0 

7' 

3 

6  4 

1   7 

8  17  41 

2  7 

14  9  4 

3 

7  8 

1   7i 

8  19  8 

2   71 

^^14  11  8 

3 

10  0 

1   71 

9  2  0 

2  71 

14  14  0 

7_^ 

3 

12  4 

1  71 

9  4  4 

2  71- 

14  16  4 

8* 

3 

14  8 

1  8 

9  6  8 

2  8 

14  18  8 

^T 

3 

17  0 

1  ^4 

9  9  0 

2  81 

15   1  0 

8i 

3 

19  4 

1  8l 

9  114 

2  81 

15  3  4 

S| 

4 

1  8 

1  sl 

9  13  8 

2  8| 

15  5  8 

9* 

4 

4  0 

1   9 

9  16  0 

2  9 

15  8  0 

^T 

4 

t)  4 

1  91 

9  18  '1 

2   91 

15  10  4 

91 

4 

8  8 

^   91 

10  0  8 

2^  91 

15  12  8 

9-? 

4 

11  0 

1   9l 

10-  3  0 

2  9^ 

15  15  0 

10 

4 

13  4 

1   10 

10  5  4 

2  10 

15  17  4J 

^^i 

4 

15  8 

I  10^ 

10  ^  l^ 

-  10  j 

15  Id   8 

lOi 

4 

18  0 

1  101 

10  10  0 

2  lOJ 

10  2  C 

lOf 

5 

0  4 

1  10^ 

10  12  A 

2  101 

]G     4   4 

11 

5 

2  8 

1  11 

10  14  C 

2  11 

2  11^ 

16  6  q 

^H 

5 

5  0 

1  Hi 

10  17  0 

16   9  0 

111 

5 

7  4 

1  ni 

10  19  4 

2  111 

16  11  4 

111 

5 

9  8 

1  111 

il   18 

2  111 

]6   13  8 

12' 

5 

12  0 

2  0 

114  0 

3  0 

16  16   0 

404 


TABLES. 


A  Comparison  of  the  American  Foot  with  the  Feet  of  other  Cininiries. 


The  American  Foot  bei 
inches,  the  feet  of  several 

America,       -     -     -     - 
London,     -      -     -     - 
Antwerp,     -     -     -     - 
Bologna,     -     -     -     - 
Bremen,       -     -     -     - 
Cologne,  -     -     -     -     . 
Copenhagen,     -     -     - 
Amsterdam,    -     -     - 
Dantzick,     -     -     -     - 
Dort,      -       -     -     -     - 
Frankfort  on  the  Main, 
The  Greek,     -     -     - 
Lorrain,     -     -     •     -     - 

Mantua, 

Meckliri,    -     -     -     -     . 
Middleburg,       -     -     - 
France,     -     -     -     -     - 
Prague,        -     .     -     . 
Rhyneland  or  Leyden, 
Kiga,    ....-, 

Koman, 

Old  Roman,     -     -     - 
Scotch,     -     -     -     -     . 
Strasburgh,     -     -     .     . 
Toledo,        -     -     .     - 

Turin, 

Venice,        -     .     -     - 


ng  divided  into  1000   parts,    or  into 
other  countries  will  be  as  follow. , 


12 


Parts. 

1000 

1000 

946 

1204 

964 

954 

965 

942 

944 

1184 

948 

1007 

958 

1569 

919 

991 

1066 

1026 

1033 

1831 

967 

970 

1005 

920 

899 

1062 

1162 


t-    - 


Inch.  Dec. 
12- 
12' 

11-352 
14-448 
11  568 
11  448 
11-580 

11  304 
ll-32a 
14-208 
11-376 
12084 
11-496 
18828 
11028 
11-892 

12  792 
12-312 
12-39G 
21-972 

1 1  -604 
11-640 

12  060 
11040 
10788 
12-744 
13-944 


A  Table  representing  the  Conformity  of  the  weights  of  the  principal 
trading  Cities  of  Europe  with  those  of  America. 

lb. 
100  of  England,  Scotland  and  Ireland, 
100  of  Amsterdam,  Faris,  Bordeaux,  k.c. 
loo  of  Antwerp,  or  Brabant, 
100  of  Rouen,  the  Viscount}',        .  - 
100  of  Lyons,  the  city, 
100  of  Hochelle, 

100  of  Thoulouse,  and  Upper  Languedoc, 
100  of  Marseilles  and  Provence, 
100  of  Geneva,         -         -         .         - 
100  of  Hamburg,  -         .         ,         . 

100  of  Frankfort, 

100  of  Leipsick,         -  -      .  -         -         - 
100  of  Bremen,  -         '         -         - 

100  of  Russia,         -  -         " 

100  of  Vienna  and  Trieste, 


of  America. 

lOOlb.  Ogz, 

109 

8 

104 

^ 

113 

14 

94 

3 

110 

9 

92 

G 

88 

11 

123 

107 

4 

111 

11 

104 

5 

110 

0 

88 

4 

123 

0 

TABLES. 


405.. 


'A  Table  representwg  the  Conformity  of  the  Weights  of  the   -principai 
trading  Cities  of  Europe  with  those  of  .America. 

lb,  '  cf  America. 


100  of  Genoa,        .        -        -         -         -         . 

=  73 

100  of  Leghorn,          -         -         -         -         . 

=  77     4 

100  of  Milan,          .,---- 

=  G3     3 

100  of  Venice,           -         -         - 

=  65   11 

100  of  Naples,       .         -         -         -         - 

=  ei  10 

100  of  Seville,  Cadiz,  &c.           -         -         . 

=  103     2 

100  of  Portugal, 

=  77     4 

100  of  Liege,             .         -         -         -         - 

=  104 

loo  of  Spain,         ---..- 

=  97 

Note,  The  Spanish  Arrobe  is  25  Spanish  pounds, 

=  24     4 

A  Table  to  cast  up  wages,  orex-| 

A  Table  to  iiud  wages  or  expense--: 

penses 

,  for  a  year,    at  so  much| 

for  a  month,  week 

or  day,  at  ec' 

per  da 

y,  week,  or  month. 
week.^bi/ month j    hy'ytar 

d. 

much  by  the  year. 

> 

Dai/.\b]/ 

by  vyrj  by  month.  |  by 

week.     1    A?.'   ■    •    j 

s.  d. 

£ 

s.  d. 

jC.  s.  d. 

jt:.  s. 

£. 

JC.     s. 

d. 

£. 

s. 

d. 

f , 

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40G 


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CO  C?  CO  CO  OO  CC  CO  CO  CO  00  CO  CO  CO  CO  CO 

4-  CO  CO  u*  c^  u,  CO  U)  w  U)  CJ  ^s  io  *a  t3 
O  <^  CO  --1  O  tn  tJ-^  Oi  tS  >—  O  CO  CO  ~^  O) 


t4 


^    f _  K 


•^  "!"  "^ 


CO  CO 


CO  CO  CO 

*o  >^  *o 

W   <0    H- 


t?     S5     J1 


TABLES, 


40^ 


AT 

ABLE 

for  reducing  Troy  wt. 

a' 

rABLE  for  reducing 

Avoirdudois  wt,  i 

to 

Avoirdupois. 

into  Troy. 

TVT 

Av. 

Troyl  Avoirdupois. 

Jv 

!           Troy, 

Jlvoir. 

1 

Troy. 

gr. 

drm. 

oz. 

lb.     oz.    dr. 

cH 

ib.  oz.  pw.    gr. 

ib. 

lb. 

oz.  pw.  gr. 

1 

•04 

1 

1      1-55 

i 

13-67 

1 

1 

2  11   16! 

2 

•07 

2 

2    3-11 

2 

3 

0  20-51 

2 

2 

5     3     8| 

-3 

4 
5 

•11 
•15 
•18 

3 

4 
5 

3  4-66 

4  6-22 

♦    5     7-77 

4 
1 

2 

1  334 

2  6  68 

o 
4 

3 

4 

7   15     0 
10     6   16 

6 

•22 

6 

6     9-32 

3 

3  10  02 

5 

6 

0  18     8 

7 

•26 

7 

7  10-89 

4 

5  1336 

6 

7 

3  10     0 

8 

•29 

8 

8  12-44 

5 

6  16-7 

7 

8 

6     1    16 

9 
10 

•33 
•36 

9 
10 

9  14 
10  15-56 

6 

7  20  04 

8 

9 

8  13     8 

11 

•40 

11 

22     1-09 

7 

8  23-38 

9 

10 

11     5     0 

12 

•44 

lb. 

8 

9     272 

10 

12 

1   16  16 

13 

•47 

1 

0  13    2-65 

9 

10     606 

.    20 

24 

3  13     8 

14 
15 
16 

•51 
'55 
•58 

2 
3 
4 

1  10     5-3 

2  7     8 

3  4  10-6 

10 
U 

11  9-4 

12  12  74 

30 
40 

36 

48 

6  10     0 

7  6  16 

17 

•62 

5 

4     1   13-25 

12 

13   1508 

50 

60 

9     3     8 

18 

'G6 

6 

4  14  15-9 

13 

14   19-42 

60 

72 

11     0     0 

10 

•69 

7 

5  12    2-56 

14 

15  22  76 

70 

85 

0   16   16 

20 
21 

•73 

•77 

8 
9 

6  9     5-21 

7  6     7-86 

15 

17     2  1 

80 

97 

2   13     8 

22 

•80 

10 

8     3  10-52 

oz 

90 

109 

4   10     0 

23 

•84 

20 

16     7     5-03 

1 

0  18     65 

100 

121 

6     6   16 

pw. 

30 

24  10  15-54 

2 

1   16   11 

200 

243 

0  13     8 

1 

3 

0-88 
1-75 
2.63 

40 
50 
60 

32  14  10-05 

41     2    4-57 
49     5  15-08 

3 
4 

2  14  16-5 

3  12  22 

300 
400 

364 
486 

7     0     0 
1     6    16 

4 

3-51 

70 

57     9     9-6 

5 

4  11     3  5 

500 

607 

7   13     8 

5 

4-39 

80 

65  13    4-11 

■  6 

5     9     9 

600 

729 

2     0     0 

c 

5-27 

90 

74    0  13-62 

7 

6     7  14-5 

700 

850 

8     6   16 

7 
8 

6-14 

7-02 

100 

200 

82    4    9-15 
164     9     2-28 

8 

7     5  20 

800 

972 

2   13     8 

9 

7-9 

300 

246  13  11-42 

'  9 

8     4     1-5 

900 

1093 

9     0     0 

10 

8-78 

400 

329     2    4-57 

10 

9     2     7 

1000 

1215 

3     6   16 

11 

9-65 

500 

411     6  13-71 

11 

10     0  12  5 

2000 

2430 

6   13     8 

12 
13 
14 

10-53 
11-41 

12-29 

600 
700 
800 

493  11     6-85 
576     0    0 
658     4     9-14 

12 

13 

10  18  18 

11  16  235 

3000 
4000 

3645 
4861 

10     0     0 

1     6   16 

15 

13-16 

900 

740    9    2-28 

14 

I  0   15     6 

5000 

6076 

4   13     8 

16 

14-04 

1000 

822  13  11-42 

15 

I  1  13  10-5   16000 

7291 

8     0     0 

17 

14-92 

2000 

1645  11     6-84 

I                         • 

18 

15-79 

3000 

2528     9     2-26 

19 

16-67 

4000 

|3291     6  13-68 

iO« 


TABLES. 


I.     TABLE 

OF  THE  MONEY  OF  COMxMERCIAL  COUx^TRIES,  WITH   THEIR 
VALUE  IN  STERLING  AND  FEDERAL  MONEY. 

MONEY  is  of  two  kinds,  Real,  and  Imaginary,  or  coin  and  money 
of  account. 

Real  Money  is  (he  coin  ofa  country,  ivhose  value  is  established 
at  a  certain  amount. 

Lnagi'nary  money  is  merely  a  denomination  to  express  a  certain 
sum»  which  is  not  represeiited  by  any  coin,  as  a  pound,  a  livre,  a 
mill.  The  denominations  of  imaginary  money  are  employed  with 
or  without  those  of  real  money  in  keeping  accounts. 

UNITED  STATES. 


Sterling:. 

£. 

s. 

d. 

Dolls.  Cts: 

Eagle,  gold  coin, 

2 

6 

0* 

10     00 

Half  Eagle,  do. 

1 

2 

6 

5     00 

Quarter  do.   do. 

0 

11 

3 

2     50 

Dollar,  silver, 

0 

4 

6 

1     00 

Half  dollar,  do. 

0 

2 

3 

0     50 

Quarter  doll.  do. 

0 

1 

H 

0     25 

Dime,               do. 

0 

0 

^ 

0     10 

Halfdime,        do. 

0 

0 

2tV 

0     05 

Cent,  copper. 

0 

0 

OH 

0     01 

Half  cent,  do. 

0 

0 

Ot^^ 

0     00-^ 

ENGLAND  AND  SCOTLAND. 

£• 

s. 

d. 

$    Cts.  Dec. 

Pound}  =20  shillings, 

1 

Q 

0 

4  44 -441- 

Slulling=12  pence, 

0 

1 

0 

0  2222f 

Fenny  =  1  farthings, 

0 

1 

1 

0  01 -So/^ 

9  pence, 

0 

0 

9 

0  16-661 

6  pence, 

0 

0 

6 

0  11.11^ 

4  pence  or  1  groat, 

0 

0 

4 

0  07  40^^ 

3  pence, 

0 

0 

3 

0  05  55| 

2  pence, 

0 

0 

2 

0  03  70t} 

6^  pence, 

0 

0 

h 

0  12  50 

3}  pencef, 

0 

0 

H 

0  06-25 

Crown,  silver. 

0 

5 

0 

1  11. n^ 

Guinea,  gold. 

1 

1 

0 

4  66'66?r 

he  English  crown  passe 

s  in  the 

U.S.  at,  <i\ 

10. 

*  Estimating  the  dollar  at  four  shillings  and  six  pence  sterling,  this  is  the  ster 
iing  value  of  the  Eagfle  ;  but  the  value  of  the  Eagle  in  English  gold  is  only 
£2  3s.  0|a.  or  5i\^>|l."stcrling. 

t  The  pound  was  anciently  a  pound  Troy  of  silver ;  it  now  contains  only  one - 
third  as  much.  The  pound  sterling  in  the  time  of  William  the  Conqueror  or 
1066,  A.  D.  was  to  the  present  pound  sterling  as  31  to  10.  Articles  were  then 
about  ten  times  cheaper  than  at  present.  So  that  his  revenue  of  ^400,000 
sterling,  was  equal  to  about  J£  12,400,000  sterling  at  present. 


TABLES.  40S 

IRELAND. 
Denominations  are  the  same  as  in  England,  but  sterling  is  to 
Irish  mpne^  as  13  to  12,  or  £12  sterling=£13  Irish. 

£.    s.    d.  D.  c.    dec. 

Pound=20  shillings,                 0  18     5f^  4   10-25f»JL 

Shiliing=12  pence,                 0     0  11-j-V  0  20'6U— 

13  pence,                                  0     10  0  22  22f 

22s.  9  pence=Eng.  guinea,     110  4  66-66^ 

4s.  IGJd.                                   0     4     6  1  00 

PRANCE. 

£.    s.    d.  D.   c.    dec. 

Livre=20  sols,                         0     0  K)  0  18-51|| 

Sol=12  deniers,                       0     0    Oi  0  0O-92if 

Denier,                                      0     0     0^\  0  00  07y%-{- 

Crown  or  Ecu=:6  livres,         0     5     0  1  IMl^ 

Pistole=*lo  livres,                    0     8     4  1851814 

Louis  d'or=24  do.                     10     0  4  44-44f 

Ecu  of  exchange=60  sols,      0     2     6  0  55.65 
The  livre  or  livre  tournois  is  estinoated  in  the  ?   ^  ini 

United  States,  at  5   "  ^"^ 

'^'  NEW  COINS. 

£.    s.    d.  D.  c.    dec. 

Franc=fi  livre  tournois,         0    0  lOj  0  18-75— 

Deciin=yV  franc,                     0     0     l^V  0  01-871— 

Centim=y^7  franc,                  0    0     Oj\\  0  00  18|— 

Five  franc  piece,  silver,          0    4     2|  0  9375 — 

2  francs,                               '      0   - 1     81  0  37-50— 

Crown  of  exchang=3  livres,  0     2     6  0  65  55| 
In  the  United  States  the  five  franc  pieces  are  }  ^  93.30 

estimated  at  y 

And  the  franc  at  0  18-731- 

There  are  silver  coins  of  the  value  of  1  franc,  and  off,  i,  and 
\  franc.     The  franc  is  to  weigh  18yYoV  g^^.  composed  of—  of  pure 

silver,  and  yV  alloy.     The  gold  coins   are  also  to  contain  j\  of 
alloy. 

£.  s.    d.  D.  c. 

20  francs,  gold,                          0  16   lOi  3  75 

40     do.      do.                             1    13     9  7  60 


SPAIN. 

Accounts  are  kept  in  money  ofvellon  or  current  dollars,  and  mo- 

aey  of  plate,  or  hard  or  plate  dollars. 

£.    s.    d. 

,      D.  c. 

Dollar,  plate=10  rials  plate,    0     4     6 

1  00 

Doll,  current  or  piastre=8  do.  0     3     7} 

0  80 

Pistanne=2  rials  plate,            0     0  lOf 

0  20 

Rial,  plate=34  maravedies,    0     0     0| 

0  10 

Quarto=4              do.               0    0     5fj 

0  01-1714. 

D  3 

4iO  TABLES 

Sterling, 

Ma  rave  di, 

100  rials  plate=188y\  rials  vel. 

100  rials  vellon=53|  plate, 

Rial  vellon, 

32  rials  vellon= 17  rials  plate. 

Ducat  of  Excbange=375  mar. 

Pistole— 36  rials  plate, 

In  the  United  States,  the  rial  vellon  is  estimated  at  5  cents,  or 
half  a  rial  plate,  and  equal  to  20  to  a  Spanish  Dollar. 

The  lower  denominations  are  different  in  some  parts  of  Spain. 
Thus,  accounts  are  sometimes  kept  in  the  following  denominations 
at 


£ 

s. 

^  a. 

$ 

c. 

0 

0 

0/t\ 

0 

00-29^^ 

2 

6 

0 

10 

0000 

1 

3 

m 

5 

31-25 

0 

0 

^u 

0 

05-311 

0 

4 

IHI 

1 

10-29,^ 

0 

16 

.  2|. 

3 

60- 

£    s. 

d. 

$    c. 

Plate  dollar=20  rials  vellon,        0     4 

6 

1   00 

Rial  veUon=8^  quartos,                0     0 

2tV 

0  05 

Quarto,                                           0    0 

OH 

0  00-58|f 

BARCELONA,  VALENCIA,  SARAGOSSA. 

£    s. 

d. 

$     c. 

Plate  dollar=37^  sols,                   0     4 

6 

1  00 

Livre=:20  sols,  "                           0     2 

H 

0  53i 

Sol=12  deniers,                           0    0 

m 

0  02| 

Denier,                                         0    0 

O/i 

0  OOf 

BILBOA. 

£    s. 

d. 

$     c. 

Plate  dollar=i20  rials  vellon,         0     4 

6 

1  0000 

Rial  vellon=-34  maravedies,         0     0 

^h 

0  05-00 

Maravedie,                                     0    0 

OfiV 

0  0014jf. 

ST.  LUCAR. 

£    s. 

d. 

$    e. 

Plate  dollar=ilO  rials  plate,          0     4 

6 

1   0000 

Rial  plate=»16  quartos,                  Q     0 

5A 

0  1000 

Q,uarto,                                           0    0 

on 

0  00-62^ 

PORTUGAL. 

£    s. 

d. 

$    c. 

Millrea=1000  reas==10testoons,0     5 

n 

1   26-00 

Rea,                                                0    0 

OtV« 

0  00-12,1 

Vintin=20  reas,                             0    0 

h\ 

0  02-60'' 

Testoon=5  vintins=100reas,      0     0 

6f 

0  1260 

Crusade  of  exchange =4  testoons,  0     2 

3 

0  60-00 

New  crusade=4|  do.                     0     2 

H 

0  60-00 

Moidore=48  testoons,                    1     7 

0 

6  00 -00 

Joanese  or  I  Johan.=64  testoons,  1    16     0  8  00-00 

The  millrea  is  126  cents  in  the  United  States. 


TABLES, 


411 


HOLLAND. 


Sterling. 


Guilder  or  Flojln=20  stivers, 

Stiver=2  grotes, 

Grote=8  pennlngs, 

Penning, 

Shilling=6  stivers=12  grotes 

Pound  Flemish=20s  =6  guilders,0  10 

Kix  dolIar=2i  guilders,  0    4 

Ducat,  1  16 

Sovereign,  1     7     ^ 

The  florin  is  40  cents  in  the  United  States. 


d. 

^1  0 

6 
0 
0 


c. 

4000 
0200 
01-00 
00- 12  J 
12-00 
40-00 
0000 
00-00 
00-00 


HAMBURGH. 

£     s. 

Pound=20  shillings  Flemish,       Oil 

Fl.  shil.=12  grotes  or  pence  Fl.  0     0 

Grote  orpenny=6  deniers,  0     0 

Mark  banco=32  grotes  or  2f  shil.  0     1 
'     Rix  donar=3  marks,  0    4 

Stiver  or  shilling  lubs,-  0    0 

Ducat=6-*  marks,  0    9 

The  Bank  money  of  Hamburgh  is  superior  to  the  currency ; 
the  agiOf  or  rate,  varies,  and  is  sometimes  20  per  cent,  or  more 
in  favor  of  the  Bank  money. 


d. 
3 

6^ 

6 
6 

H 


$    c. 
2  60 
0  12-600 
0  01^ 

0  33i 

1  00 

0  02^V 

2  16| 


BREMEN. 

£ 
Rix  dollar=72  grotes=2i  marks,  0 
Grote,  0 


d. 


0^ 


$  c. 
0  76 
0  01^,- 


ANTWERP. 

£  s.  d. 

Guilder=3i  shillings=:40  grotes,    0  1  9f 

Shilling=12  grotes,                          0  0  6^f 

Grote,                                                0  0  0|^ 


Stiver=^2  grotes, 


0    0 


l/« 


$  c. 
0  40 
0  12 
0  01 
0  02 


VIENNA  AND  TRIESTE. 


Florin=60  cruitzers, 
Cruitzer=i=4  fenings, 
Batzen=4  cruitzers, 
Rix  dollar: 
Livre=20  soldi, 
Specie  dollar=30  batzen, 
Ducat=60  batzen, 


li  florins, 


d. 
4 

HI 


H 


61-85/^ 


$ 
0 

0  00-866— 

0  03-46 

0 

0 

1 

2 


m 

09-87 
03  7 
07-4 


The  value  of  these  denominations  is  the  same  generally  through 
Austria,  Swabia,  Frauconia,  Bohemia,  Silesia,  and  Hungary. 


412  TABLES. 

HANOVER  AND  SAXONY. 

Sterling. 

£s.  d.  $    c. 

Rix  dolIar=24  groshen,  0  3  6  0  77j 

Grosh=12  fening,  0  0  If  0  03if 

Gould  orguilder=16groshen,  0  2  4  0  61-85^^ 

Ducat=4  guilders  or  goulds,  0  9  4  2  07-40 

Specie  dollar  or  hard  dollar,  0  4  8  1  03-7 

POLAND  AND  PRUSSIA. 

£     s.  d.  $   c. 

His  dollar=90  groshen,  0     3  6  0  77f 

Florin       =30     do.  0.    1  2  0  25^ 

Grosh=3  shelons,  0    0  Oj\  0  00-866- 

Ducat=8  florins,  0     9  4  2  07-4 

Frederic  d'or=5  rix  dollars,  0  17  6  3  88f 

LIVONIA. 

£  s.  d.  $    c. 

Rix  dollar==90  groshen,  0  3  6  0  77f 

Florin       =30     do.  0  12  0  25ff 

Marc         =  6     do.  0  0  2^  0  05/^ 

Grosh=6  blackens,  0  0  0/j  0  00-866"- 

RUSSIA. 

JE   s.  d.  $   c. 

Ruble=100  copec5,  04  6  100 

Poltin=  60     do.  0  2  3  0  60 

Copec=4  poluscas,  0  0  0|4  0  01 

2ervenitz=2  rubles,  0  9  0  2  00 

DENMARK  AND  NORWAY. 

£  s,  a.  $  c. 

Rix  dollar=6  marks,  0  4  6  1  00 

Crown      =4     do.  0  3  0  0  66| 

Marc=16skilling8,  0  0  9  0  16f 

Skilling,  0  0  OyO^  0  01  «f 

Ducat=^ll  marcs,  ,  0  8  3  1  83| 

SWEDEN  AND  LAPLAND. 

£  s.  d.  $    c. 

Rix  dollar=3  silver  dollars,  0  4  8  1  03^^ 

Silver  doUar=3  copper  dollars,  0  1  6|  0  34|f 

Copper  dollar=4  copper  marcs,  0  0  6|  0  1  l-}f | 

Copper  marc=8  runstics,  0  0  ]|  0  02|-ff 

Ducat=2  rix  dollars,  0  9  4  2  07^-} 

SWITZERLAND. 

£  s.  d.  $    c. 

Rix  dollar=108  cruitzers,  0  4  6  l  00 

CruitzeT=4  feaingf,  0  0  0}  0  00S>2i^ 


TABLES. 


413 


Sterling. 

£s,  d. 

$    c. 

Fening=3  rap, 

0  0  Oi 

0  oo-ss^v 

Sol=12f€nings, 

0  0   li 

0  02-771 

Livre  or  gulden=20 

sols, 

0  2  6 

0  65f 

GENOA,  CORSICA,  &c. 

£     s.   d. 

$  <!. 

Pezzo  of  exchange= 

=  115  soldi, 

0     4  2 

0  92-59/^ 

Soldi=12  denari, 

0  0  o^ 

0  00-80 

Lire=20  soldi, 

0     0  8if 

0   16-10 

Testoon=30  soldi, 

0     1  Oi| 

0  24-1 S 

Pistole =20  lires, 

0  14  6^V 

3  22-22 

SARDLNIA, 

TURIN,  AND  SAVOY, 

£     s.  d. 

$   ^' 

Scudi=6  florins, 

0     4  6 

1  00 

Florin=12  soldi, 

0     0  9 

0  162 

Soldi=:12  denari, 

0     0  01 

OOItV 

Lire=20  soldi, 

0     1  3 

0  27|- 

Ducatoon=7  florins 

> 

0     6  3 

1    16| 

Pistole=13  lires. 

0  16  3 

3  61i 

Louis  d'or=16  lires 

* 

1  00  0 

4  44f 

ROME  AND  CIVITA  VFXCHIA. 

riurrent  crown=10 

julios 

0     6  0 

1  ni 

Julio=8bayocs 

0     0  6 

0   11* 

Bayoc=5  quatrini 

0     0  03 

0  oiVv 

Stamped  Julio=10  1 

bayocs 

0     0  71 

0  13| 

Stamped  crown=12 

julios 

0     6  0 

1  33* 

Pistole=31  julios 

0  15  6 

3  34^ 

FLORENCE  AND  LEGHORN. 

Piafitre=6  lires 

0     4  0 

©  88| 

Lire =20  soldi 

0     0  8 

0  14-81-i 

Soldi=12  denari 

0     0  Of 

0  01-23 

Ducat=7i  lires 

0     5  0 

1    Hi 

Pistole  =22  lires 

NAPLES. 

0   14  S 

3  25-93 

Pucat=100  grains 

0     3  4 

0  74-07|i 

Orain=3  quatrini 

0     0  0/^ 

0  00-74 

Carlin=10  grains. 

0     0  4 

0  07-40 

Ounce=3  ducats 

0   10  0 

2  22-225^ 

Pistole =4-6  carl  ins 

VENICE. 

0  15  4 

£  s.  d. 

3  33-33^ 

Ducat=24  gros, 

0  4  4 

0  96/,. 

Gros  or  grosi=5J-  soldi, 

0  0  2J- 

0  04J^ 

SQldi=12  denari, 

0  0  O^f 

0  00-77? 

Lire=20  soldi, 

0  0  8if 

0  15-53^ 

4J4 


TABLES. 


In  Leghorn,  a  piastre=  20  soldi  of  Genoa. 
Naples,  a  ducat      =  86     "  ** 

Milan,  a  crown       =  80     "  " 

Sicily,  a  crown       =127|  "  " 


PALERMO  IN  SICILY. 

Sterling. 

£  s.      d. 

Once  or  Onge=30  tarins,  0  10 

Tarin=20  grains,  0  0 

Grain,  0     0 

Dollar  of  Sicily=240grs.=12  tarin,  0     4 

Spanish  dollar=252  grains,  0  4 


9f 
0-2JL 

^125 
*^2  5 


6 


$     C. 

2  40 
0  08 
0  00*4 

0  96 

1  00 


TURKEY. 


Piastre=80  aspers, 
Asper=4  inangars, 
Parac=3  aspers, 
Ostic: 

Caragrouch=100  aspers, 
Xerifi=10  solotas=200  aspers, 


10  do.=i  solota, 


£    s. 
0     4 


0  0  Of 

0  0  If 

0  0  6 

0  6  0 

0  10  0 


0  88f 
0 
0 
0 

1 
2 


03'33i 
IMli. 

mix 

22-22| 


Piastre=100  aspers  or  40  paras, 

Asper, 

136  Paras=Spanisb  dollar, 

Para, 

10  Spanish  dollars=34  piastres. 


SMYRNA. 

£ 

0 

0 

0 

0 


ARABIA. 

Piastre=60  comashees=80  caveers, 
Comasbee=7  carrets, 
Larin— 80  do. 
SeqMin=100  comashees, 
Tomond=B0  Larins, 


d. 
3H 


on 


$   c. 

0  29'41y\ 

0  00-29 

1  00  00 

0  00-73/^ 
10  00 


d. 
6 
0-^- 

'^1  0 

lOf 
6 
6 


$  c. 
1  00 
0  01 '661 

0  1905 

1  66  66f 
15  00 


EGYPT. 


PiastFe==80  aspers, 

Asper, 

Medin=3  aspers, 

Dollar=^30  medins, 

Ecu  or  crown=96  aspers, 

Sultanin=2  ecu, 

?argo  dollar=70  medini, 


s. 
4 
0 
0 
4 
5 
10 
10 


d. 
0 

If 

a 

0 
0 
6 


OlllJ^ 
03-33/^ 


00 
11 

22-22f 
33-33 


m 


TABLES. 


415 


ALGIERS,  TUNIS  AND  TRIPOLI. 


Sterling. 

£    s.     d. 

$  c. 

Dollar=:4  doubles. 

0     4     6 

1   00 

DoubIe=2  rials, 

0   1   n 

0  25 

Rial=10  aspers, 

0     0     6| 

0  12.', 

Pistole=^16  doubles. 

0  16   10^ 

2  75 

Zequin=180  aspers, 

0  10     1^ 

2  25 

Pataca  Chica, 

0     0   lU 

0  21/. 

Sultanin=8i  patacas  chicas, 

0     8      If 

1   81 

Piastre=3         "             " 

0     2   lOi 

0  64 

MOROCCO,  FEZ, 

TANGIERS,  AND  SALLEE. 

£    s.     d. 

$    c. 

Dollar=28  blanquils, 

0     4     6 

1  00 

Blanquil=24  fluces, 

0     0     1-H 

0  03-57! 

Ounce=4  blanquils, 

0     0     7f 

0  14-28-i 

Q,uarto=14     '* 

0     2     3 

0  50 

Zequin=56     " 

0     9     0 

2  00 

Pistole =100  " 

0  16     04 
PERSIA. 

3  57| 

£     s.    d. 

$   c. 

Bovello=12  abashees, 

0   16     0 

3  55-56f 

Abashee=4  shahees, 

0      1      4 

0  29  63— 

Sbahee=10  coz, 

0     0     4 

0  07-41-~ 

Larin    =25  do. 

0     0  10 

0  18-51 

Or=5  abashees, 

0     6     8 
BOMBAY. 

1  4314 

£     s.     d. 

$    c. 

Rupee=4  quarters, 

0     2     3 

0  50 

Quarter— 20  pices=100  reas 

,                   0     0     6f 

0  m- 

Rea, 

0    0    o^v^ 

0  00-125 

Pagoda=14  quarters, 

0     7  lOi 

1   75 

Rupee,  gold=GO  do. 

1    13     9 

7  50 

Current  niohuss=15  current 

rupees,     1     1     1| 

4  68f 

Current  rupee=50  pice. 

0     1     41 

0  31{ 

MADRAS  AND  PONDICHERRY. 

£    s.   d. 

$    c. 

Pagoda=S6  fanauis, 

0     8  3/^ 

1  84 

Fanam=8pice=80  cash, 

0     0  2f 

0  05i 

Pice=2  viz=lO  cash, 

0   0  014 

0  00  64— 

Rupee  =  10  fanams, 

0     2  3f 

0  51J- 

Crown=2  rupees. 

0     4  7^ 

1   02f- 

Rupee,*  gold=4  pagodas, 

1    13  1|4- 

7  36 

Bengal,  or  new  Sicca  rupee, 

0     2  3 

0  50 

*  Lack  of  rupees  is  100000,  and  340  Sicca  rupees  pas3  currently  f6r  100  Star 
pagodas, 


416  TABLES. 

Pagoda  is  184  cents  in  the  United  States. 
Sea  shells,  called  cowries  are  used  for  change  ;  their  value  va- 
rigs  with  their  quality. 

CALCUTTA  AND  CALLICUT. 

Sterling. 

£  s.  d.  $  c. 

Rnpee  =  16  annas,                                      0  2     3  0  60 

Anna=12  pices,                                           0  0     1||  0  03-|- 

Fanam=4     do.                                           0  0     6f  0  12| 

Crown  or  Ecu=2  rupees,                         04     6  1  00 

Pagoda=56  annas,                                      0  7  10^  1  75 
100  Sicca  rupe€S=116  current  rupees. 

PEGU,  JAVA,  SUMATRA,  fcc. 

£  s.  d.  $    c, 

Dollar=900  fettees,  0  4  6  1  00 

Fettee=10  cori,  0  0  0/^  0  OOJ- 

Tical=500  fettees,  0  2  6  0  55f 

Rial,  crown  or  ecu=2  ticals,  05  0  1   1^ 


CHINA. 


JE  s.  d.  $ 


c. 


Tale=10  maces,  0  6  8  1  48^^ 

Mace=10  candereens,  0  0  8  0  14"81 

Candereen=10  cash,  0  0  0|  0  01-48 

The  Spanish  dollar  passes  at  72  candereens,  which 

makes  the  candereen  worth,  0  01-38| 

But  the  tale  is  reckoned  in  the  United  States,  1  4800 


JAPAN. 


£  s.  d. 

$    c. 

Tale=10  mace=rix  dollar, 

0  3  4i 

0  75 

Mace=10  candereens. 

0  0  4^V 

0  07i- 

Candereen=2  pitis, 

MANILLA. 

0  0  2,V 

0  03-7^ 

£  s.    d. 

$    c. 

Dollar~8  reals. 

0  4  6 

1  00 

Real=12  quartos, 

0  0  6^ 

0  m 

quarto, 

BATAVIA. 

0  0  Oy\ 

0  oC\ 

£  s.  d. 

$    c. 

Spanish  dollar=64  stivers. 

0  4  6 

1  00 

Stiver, 

0  0  op 

0  OlVk 

Uix  dollar=48  stivers 

0  3  4f 

0  75 

Ducatoon— 80  do. 

0  5  7X 

1   25 

TABLES.  417 

COLOMBO  IN  CEYLON. 

je  s.  d.  $    c. 

Spanish  dolIar=64i  stivers,  0  4  6  1  00 

Sliver,  0  0  Of-«  0  01-55 

Rupee=30  stivers,  0  2   I/3  0  46'63i 

Rix  doIIaf=48  do.=8  shillings,  0  3  4^\  0  74-45 


ENGLISH  WEST  INDIES. 

The  principal  difference  is  in  the  number  of  shillings  in  the 
Spanish  dollar,  while  a  pound  is  20  shilling's,  and  a  shilling  is  12 
pence,  at  Jamaica  and  Bermudas,  the  Spanish  dollar  is  6  shillings 
and  8  pence.     Hence  the 

£   s.    d.  $    c. 

Pound=20  shillings=3  dollars,  0  13  6  3  00 

Shining=12  pence,  0     0  Sj\         0   15 

Penny,  0     0  0||         0  01-25 

At  Barbadoes,  the  Spanish  dollar  is  6  shillings  and  3  pence. 
Hence  a 

Pound=20  shillings  0  14  4f  3  20 

Shilling=12  pence  0     0  8i|         0  IG 

Penny  0     0  Off         0  01^ 

FRENCH  WEST  INDIES. 
In  some  ofthe  Islands  the  Spanish   dollar  passes  for  8  livres  and 
6  sols,  and  in  others  for  9  livres.     In  the  former,  the 

Livre=20  sols 
Sol=12  deniers 

8  livres  5  sols 
In  the  latter,  the 

Livre=20  sols 
Sol=12  deniers 

9  livres 

IN  MARTINICO,  TOBAGO  AND  ST.  CHRISTOPHERS, 

The  English  inhabitants  keep  their  accounts  in  the  denomina- 
tions of  English  money,  and  the  French,  in  those  of  France.  But 
the  round  dollar  passes  for  9  shillings,  and  the  current  dollar  at  8 
shillings  and  three  pence,  or,  the  round  is  to  the  current  dollar  as 
12  to  11.  So  that  99  livres=ll  round  dollars  or  =12  current 
dollars. 


0 
0 
0 

s.  d. 

0  6A 
OOif 
4  6 

D.    c. 

0  123V 

0  00-60§f 

1  00 

0 
0 
0 

0  6 
4  6 

0   lli 

0  00-55f 

1  00 

SPANISH  WEST  INDIES. 

Dollar==8  reals 
Real 

£.    s.    d. 
0     4  6 
0     0  6£ 

$   c. 
1  00 
0  12i. 

E  3 


410 


TABLES. 


TABLE  II. 


OF  THE  MONEY  OF  THE  JEWS,  GREEKS,  AND   ROMANS,  "WITH 
THE  VALUE  IN  STERLING  AND  FEDERAL  MONEY.* 


\' 


Talenl=60  manch 
Manch,  or 
Hebrew  mina 
Shekel=2  becah 
Becah=10  gerah 
Gerah 
Sextula 
Siclus  aureus 
Talent  of  Gold 

Iti  this  estimate  the 
n  to  1. 


OF  THE  JEWS. 

je.  s. 

d. 

$   c. 

342  3 

9 

1620  83|- 

60  shekels 

6  16 

m 

30  41| 

0  2 

3f 

0  50|f 

0  1 

lU 

0  26f|- 

0  0 

Hh 

0  1/^ 

0  12 

^ 

2  67/^ 

1  16 

6 

8  111 

5475  0 

0 

25533  33i  • 

value  of  gold  is  to  that  of  silver   nearly  as 


OF  THE  GREEKS. 


Drachma— 1|  tetrobolom 
Tetrobolum=2  diobolum 
DioboIum=2  obolus 
Oboliis=2  hemiobolum 
Hemiobolum=4  chalcus 
Chalcus=7  septon 
Didrachmon=2  drachma 
Tetrard  statu=2  didrachma 
Mina=100  drachmae 
Talent=60  minaa  . 
IOC  Talents 

Statu  aureus=25  drachma 
Statu  daricus— 50  do. 
according  to  Josephus, 


s.  d. 
0  71 


£ 
0 
0 
0 
0 
0 
0 
0 
0 
3 

193  15  0 
19375     0  0 

0  16   13 

1  12  31 


0  2tV 

0  lA 
OOff 

1  31 

2  7 
4  7 


$    c. 
0  14i| 
0  09ff 
0  04if I 
0  02i|l 
0  Olifl 
0  OOlneariy 

0  281^ 

0  5711 

14  35-V 


861    111 
86111    111 
3  58f^ 

7  17if 


OF  THE  ROMANS. 


Denarius~2  quinarii 
Q,uinariu8=2  sestertii 
Sestertius=2^  asor  libella 

=4  teruncii 
Teruncios 
Sestertium=1000  sestertii 


*  Auihora  differ  respecting  the  precise  value  of  ancient  money, 
estimate  is  here  g^iven,  which  is,  at  least  sufficiently  near  the  truth. 


0 
0 

s.  d. 
0  7f 
0  3| 

$    c. 
0  14if 
0  07J^ 

0 

0  Hf 

0  03lfl 

0 

8 

0  ^i 

1 51 

0  00  j% 
35  87 

nearlj= 


The  commop 


TABLES. 


419 


d. 


D. 


8072  18  4       35879  63 


Decius  sestertium=1000  ses- 
tertia, 

Centies  sestertium,  or  centies 

HS.  was  10000  sestertia,  or  80729     3  4 

Millies  HS.  was  100000  ses- 
tertia, or  807291    13  4 

And  the  millies  centies  HS. 
was  the  sum  of  the  last 
two,  or  888020  16  8 

Aureus=2i3  denarii  0  16   If 

This  ratio  of  the  aureus  to  the  denarius  is  that  mentioned   by 

Tacitus. 


3  68H 


TABLE  III. 


OF  MEASURES  OF  LENGTH  AND  CAPACITY  AND  WEIGHTS  OF 
VARIOUS  COUNTRIES. 


THE  table  of  English  and  American  measures  has  been  given 
under  compound  addition. 

Compared  with  the  French  measure,  the  English  inch  is 
0-253994053958632382 12354.  of  a  French  metre. 


A  Foot  English 

Yard=3  feet 

Rod  or  pole=5i  yards 

Mile=320  rods 

League=3  miles 

Ell  Englishes  quarters  of  a  yard 

Ell  Flemish=3  do. 

Ell  French=6  do. 

Wine  gallon=231  cubic  inches  English 

Ale  do.=282  do. 

Gallon,  dry  measure— 268-8  do. 

Bushel=32  quarts=2150-42  do. 

Wine  quart=57*75  do. 

Ale  do.=70-5  do. 

Dry  do.=67-2  do. 

Wine  Hogshead=63  gallons 


French  metres. 

0-30479286-1- 

0-914378594* 

5-02908227— 

1609-30632588-h 

4827-919977644- 

M42973 

0-685784 

1-371568 

French  litres. 

3-3735 

4-6208 

4.4043 

.35-2343 

0-9463 

M552 

11011 

238-4509 


SCOTLAND. 


3  Feet  make  1  ell, 

1  Mile=5760  feet, 

30  Scotch  ells  are  equal  to 

30  Scotch  miles, 


English. 
37-2  inches. 
5952  feet. 

31  yards. 

31  miles. 


426 


TABLES. 


I  Fall=6  ells,  2234  inches 
48  Scotch  acres  are  very  nearly  61  acres- 

For  the  measure  of  Wheats  Peas,  Beans,  Rye,  and  While  Salt, 
100  Bolls  equal  409  bushels,  Winchester  measure. 

"For  Barley,  Oats,  and  Malt. 
100  Bolls  equal  596  bushels,  Winchester  measure. 
f{ote.  The  Boll  varies  in  d^^rent  parts  of  Scotland. 

IRELAND. 

The  Irish  and  English  foot  and  yard  are  equaL 
The  Irish  mile,  =2240  yards. 

II  miles  Irish,  14  miles. 
121  acres  "                                                            196  acres. 

The  Irish  bushel  contains  1740-8  cubic  inches. 


Metre, 

Deca-metre=10  metres, 
Hecto-metre=100  " 
Kilo-metre=1000    " 
Myria-metre=  10000  " 

Deci-metre=y^j  metre, 

Centimetre=T 

Milli-metre=- 


i 

'O  0 

J 

10  0  0 


FRANCE.* 

French  feet. 
3078444 
30-78444 
307  8444 
3078-444 
30784-44 

French  inches. 
3-6941328 
0-36941328 
0-036941328 

French  lines. 
443-295936 


English  feet- 
3-2809167 
32-809167 
328-09167 
3280-9167 
32809-167 

English  inches. 
3-93710004 
0-393710004 
0-0393710004 

English  lines. 
472-4520048 


Metre, 

Quadrant  of  the  Meridian,  100  French  degrees,  90  English  degrees. 


*  France.is  the  only  nation,  which  has  established  an  invariable  standard  of 
measure.  The  linear  unit  of  the  French  measure  is  the  metre.  By  accurate 
observations  and  calculations  the  length  of  the  meridian  from  the  Equator  to 
the  pole,  v.'hich  passes  through  the  city  of  Paris,  was  ascertained  to  be  5130740 
toises  of  six  feet  each  of  the  ancient  French  measure.  This  number  of  toises 
is  equal  to  30784440  French  feet,  or  32fl091G7  English  feet.  The  metre  is  one 
ten  millionth  part  of  this  arc  of  the  meridian,  or  3-078444  feet,  which  is  3  feet 
J 1  lines,  and  _2iL?_93_s_  of  a  line  of  tlie  former  French  measure.      All  other 

10  0  0  0  0  0 

measures  are  derived  from  the  metre. 

In  England,  it  has  been  proposed  to  make  the  length  of  the  pendulum  to  vi- 
brate seconds  at  London,  the  standard  of  measure.  At  the  level  of  the  sea,  and 
when  the  temperature  is  62d.  Far.  and  in  lat.  51d.  ?iV  8-4"  N.  the  length  of  the 
pendulum  to  vibrate  in  a  second  is  30-1386  inches,  English,  as  very  accurately 
•.ietermined  by  Capt.  Katar.  According  to  Capt.  Kater's  measure,  the  French 
metre  is  39-37076  inches  English,  at  the  same  temperature,  and  may"  be  taken 
with  sufficient  accuracy  to  be  39-371  inches.  Thc^o  measures  will  vary  a  little 
according  to  the  scale  on  which  they  are  estimated.  U  Troughton's  scale  of  36 
inches  be  taken  as  the  standard,  General  Roy^s  Scale  is  36-00036  inches,  and 
Bird's  Parliamentary  standard  of  1758,  is  36-000^3  inches.  And  if  the  scale  of 
1758  be  the  standard,  Troughton's  scale  is  35-99077  inches.  According  to  Mr 
'l£as?lcr,  the  French  metre  is  3-23168733  feet  on  Troughton's  Scale. 


French  feet. 

English  feet. 

6 

6-3946266-h 

3  63f 

3-87824. 

1- 

1  0657711+* 

Ot\ 

0'0888I42-1-. 

TABLES.  421 

French  feet.  English  feet, 

Degree=54  min.  Eng.  100  min.  Fr.         307844-4  328091-67 

Minute==32'4  sec.  Eng.  lOOsec.  Fr.         3078-444  3280-9167 

The  are  is   the   square  of  the  deca-metre,  and  is  the  unit  for 

square  measure. 

French  square  feet.      English  square  feet. 

Are,  947-681746113  107644143923 

Square  metre  or  Centiare,         9-47681746113         10-7644143923 

The  litre  is  the  cube  of  the  decimetre,  dind  is  the  unit  for  dry  and 

liquid  measure. 

French  cubic  inches.  Eng.  cubic  inches. 
Litre,  60-412416  61028 

Hecto!itre=13  veltes  3i  pints,  «  6102  8 

Paris  pint  is  yVoWoVo  ^^^^^  ^^  46-95  cubic  inches. 
The  stere  is  a  cubic  metre,  and  is  the  unit  for  Cubic  or  Solid 
Measure. 

French  cubic  feet.     English  cubic  feet. 
Stere,  20-17385  35  317 

OF  THE  OLD  FRENCH  MEASURES. 

The  Toise, 

Aune  or  Ell, 

Foot=12  inches, 

Inch=12  lines, 
A  French  League  is  nearly  2J  English  miles,  or  about  j^  of  an 
English  League. 

By  a  decree  of  1812,  the  Toise,  Aune,  Foot,  &c.  are  allowed  to 
be  the  denominations  of  measure  for  the  common  people  of  France, 
in  the  following  ratios  to  the  metre. 

Toise=2  metres,  6  56  English  feet  nearly. 

Foot—}  metre,  05468 

lnch=yV  metre,  0  2734 

Aune  or*'ElI=li  metre,  3-9371 

Bushel^i  Hectolitre,  762  854  cubic  inches. 

The  old  litron=40'393455i|^  French  cubic  inches,  by  statute,  but 
the  common  iitron  is  48  8224  English  cubic  inches. 

Weight.  .  English. 

Paris  pound  or  16oz.=:2  marcs,  7560grs. 

Ounce,  4725 

16  pounds  are  211bs.  Troy. 

63oz.  64oz.  Troy. 

100  pounds       1081bs.  Avoirdupois, 

92if  do.  100  do. 

25       do.  27     do. 

1  Kilogramme  45-35  do. 

1  Hectogramme  453.50  do. 

'■  The  ratio  of  the  French  to  the  Eu2;lish  foot  here  assigned  is  very  little  cui- 
I'orcnt  from  1  to  1  jl^g,  which  v/as  furaierly  considered  as  the  true  ratio. 


422 


TABLES. 

HAMBURGH. 

English. 

100  Ells, 

62^  yards. 

16    do. 

10      do. 

1  German  mile, 

4  miles. 

100  lbs. 

1074  lbs. 

93iM> 

100    do. 

1  Shippound, 

280     do. 

SWEDEN. 

1026275  feet, 

1000  feet. 

1000          do. 

974-397  do. 

1         Can  or  Kann, 

159-864  cub.  inch, 

877-7125  Victualievigt, 

lOOOlbs.  Troy. 

1333-4935  Skulpounds, 

1000  do.  Av. 

HOLLAND. 

100         Ells, 

75yds. 

1331         do. 

100  do. 

1         Dutch  mile. 

3i  miles. 

1         ife=2marcs=16oz. 

5775grs. 

100        lbs. 

109ilbs. 

9I2V9  <^o- 

100  do. 

219         do. 

200  do. 

ANTWERP. 

100       Brabant  Ells, 

74yds. 

135/^              do. 

100  do. 

100       lbs. 

104ilbs. 

,  96       do. 

100  do. 

Q,uintal=10  myriogrammes, 

2041  do. 

100       pots  of  Brabant, 

361  gallons. 

BREMEN. 

100       fe, 

110ft. 

90if  do. 

100  do. 

1       Last, 

80  bushels. 

DENMARK. 

96  lbs. 

lOOlbs. 

100  lbs. 

1041  do. 

1  Shippound=320]bs.  or  20  } 
lispounds,                        J 

33.31  do/ 

K^tyt^^  vlV/k 

1000  Feet, 

1049 

RUSSIA. 

1  Arsheen, 

28  inches. 

9      •  do. 

7  yards. 

TABLES. 


423 


100  lbs. 

English. 
SBiibs. 

1  Pood=40lbs. 

33,Vdo. 

1  Borquit=10  poods, 

333      do. 

Russian  Verst, 

f  mile. 

SPAIN. 

Vara  or  Barra=i|  of  French  ell, 

2-7471  feet. 

Vara,  of  Catalonia=li        do. 

5-8173    do. 

lOOlfe, 

97ft. 

Arobe=25lbs. 

241  do. 

Spanish  league, 

3|  miles. 

BILBOA. 

100  Varas, 

108yds. 

100  ft, 

106ilbs. 

100  ft  of  iron, 

100  ft. 

32  velts. 

66  gallons. 

100  fanagues, 

152  bushels. 

PORTUGAL. 

Cavedo=26|  Eng.  in.  or  accurately. 

26-6933  inches. 

Vara        43|       do.                 do. 

44-66 

32ft=l  arobe,                            nearly 

33fe. 

Q.uintal=4  arobes  or  128fe 

132ft. 

60  Alquiers  or  1  Moy, 

24    bushels. 

Fanga=4  alquiers, 

If       do. 

Canado=4  quarteels. 

3    pints. 

Almude— 12  canados, 

41  wine  gallons 

NAPLES. 

Cane,                                 =7  feet,  or 

84  inches. 

Pound  of  silk, 

12oz. 

1001  bs. 

64|lbs. 

Cantar=100  rotolos, 

196lbs. 

LEGHORN. 

145  lbs. 

n2ft. 

100  brasses, 

64yds. 

Cane=4  brasses. 

2ifyds. 

116  sacks,                       100  quarters, 

800  bushels. 

TRIESTE. 

Brace,                              27  inches,  or 

f  yard. 

100  ft  of  Vienna, 

123    ft  Av, 

31  Staros,                               1  quarter. 

,      8    bushels. 

Staro, 

2f  ■  do. 

Barrel  of  wine, 

18    gallons. 

124 


TABLES. 

PALERMO  IN  SICILY. 

Englisli. 

Cantar, 

1761bs.  Av. 

Rottoli=Yj,^  canlar, 

lllfe  nearly. 

Po'Mui  r;eight, 

fife. 

Salm. 

485  fe. 

Caffis, 

3i  gallons. 

Caular  .of  oil, 

25       do. 

Barrel, 

9       do. 

SMYRNA. 

Pike, 

1  yd.  nearly 

45  Okes,                                  nearly 

123|  ft  Av. 

40|  do. 

112     do. 

BOxMBAY. 

Maud, 

iCwt. 

Surat  maud, 

37ilbs. 

Surat  Candy^21  S.  mauds, 

784  do. 

Bombay  Candv=21  B.  mauds, 

588  do. 

The  ife  is  the  English  3fe. 

MADRAS. 

Picul=100  cattas, 

133ilbs. 

Calta, 

Hife. 

Maud, 

25fe  Troy, 

Candy=20  mauds, 

500lbs.  do. 

CALCUTTA. 

Cavid, 

^l\ 

Bazar  maud, 

fCwt. 

Factory  do. 

75  ife. 

1  Maud=40  Seer?,  and 

1  Seer=16  chittacks. 

CHINA. 

Covid, 

Uj%  inches. 

Picul=100  cattas, 

1331  lbs. 

Catta=^16  tales, 

I3-     *• 

BATAVIA. 

Picul  =  100  cattas  =  125lfe  Dutch 

=  1335ife. 

MANILLA. 

lOOfe, 

101%. 

Arobe  or  25ife, 

26  do. 

Picul=5i  arobes, 

143  do. 

ACCOUNT  OF  THE  GREGORIAN  STYLE,  &c.       425 

JAPAN. 

English. 
Hichey  or  Ichan,  3|  feet. 

Catta=16  mace,  13^fe. 

IVIace  =  10  tales,  |  do. 

Balec,  16i  gallons. 

FRENCH  WEST  INDIES. 
104ife,  112     lbs. 

100ft,  lOTy^ibs. 


AN"   ACCOUNT   OF   THK    OUKGORIAN   OR   NEW   STYLE,   TOGETHER  WITH  SOME 
CHRONOLOGICAL  PROBLEMS,  FOR    FINDING  THE    EPACT,  GOLDEN   NUMBER, 

moon's  AGE,  &:c. 

POPE  GREGORY  the  XIMth  made  a  reformation  of  the  cal- 
endar. The  Julian  calendar,  or  old  style,  had,  before  that  time, 
been  in  general  use  all  over  Europe.  The  year,  according  to  the 
Julian  calendar,  consists  of  three  hundred  and  sixty  five  days  and 
six  hours  ;  which  six  hours  being  one  fourth  part  of  a  day,  the 
common  years  consisted  of  three  hundred  and  sixty  five  days,  and 
every  fourth  year,  one  day  was  added  to  the  month  of  February, 
which  made  each  of  those.jyears  three  hundred  and  sixty  six  days, 
which  are  usually  called  leap  years. 

This  computation,  though  near  the  truth,  is  more  than  the  solar 
year  by  eleven  minutes,  which.  In  one  hundred  and  thirty  one 
years,  amounts  to  a  whole  day.  By  which  the  Vernal  iKquinox 
was  anticipated  ten  days,  from  the  time  of  the  general  council  of 
Nice,  held  in  the  year  325  of  the  Christian  iEra,  to  the  time  of 
Pope  Gregory  ;  who  therefore  caused  ten  days  to  be  taken  out  of 
the  month  of  October  in  1582,  to  make  the  /Equinox  fi\ll  on  the 
21st  of  March,  as  it  did  at  the  time  of  that  council.  And,  to  pre- 
vent the  like  variation  for  the  future,  he  ordered  that  three  days 
should  be  abated  in  every  four  hundred  years,  by  reducing  thQ 
leap  year  at  the  close  of  each  century,  for  three  successive  cen- 
turies, to  common  years,  and  retaining  the  leap  year  at  the  close 
of  each  fourth  century  only. 

This  was  at  that  time  esteemed  as  exactly  conformable  to  the 
true  solar  year;  but  Dr.  Halley  makes  the  solar  year  to  be  three 
hundred  and  sixty  five  days,  five  hours,  forty  eight  minutes,  fifty 
four  seconds,  forty  one  thirds,  twenty  seven  fourths,  and  thirty  one 
fifths:  According  to  which,  in  four  hundred  years,  the  Julian  year 
of  three  hundred  and  sixty  live  days  and  six  hours  will  exceed  the 
solar  by  three  days,  one  hour  and  i\hy  five  minutes,  v/hich  is  near 
two  hours,  so  that  in  fifty  centuries  it  will  amount  to  a  day. 

Though  the  Gregorian  calendar,  or  new  style,  had  long  been 
used  throughout  the  greatest  part  of  Europe,  it  did  not  take  place 
in  Great  Britain  and  America  till  the  first  of  January,  1752;  and 
in  September  following,  the  eleven  days  were  adjusted  by  calling 
the  third  day  of  that  month  the  fourteenth,  and  continuing  the 
rest  in  their  order. 

F  3 


126  CHRONOLOGICAL  PROBLEMS. 


CHRONOLOGICAL  PROBLEMS. 

Problem  I. 

As  there  are  three  leap  years  to  be  abated  in  every  four  centuries :  tQ 
shew  how  to  find  in  which  century  the  last  year  is  to  be  a  leap  year^ 
and  in  which  it  is  not. 

Rule. 
Cut  off  two  cyphers,  and  divide  the  rettiaining  figures  by  4 ;  if 
nothing  remain,  the  year  is  a  leap  year. 

ExAMP.  1.  The  year  18|00.  Examp.  3.  The  year  20|00. 

4)18(4  4)20(5 

16  20 

2  0 

Examp.  2.  The  year  19iOO.  Examp.  4.  The  year  40|00. , 

4)19(4  4)40(10 

16  40 

3  0 

The  first  and  second  examples,  having  remainders,  shew  the 
years  to  be  common  years  of  three  hqndred  and  sixty  five  days  ; 
but  the  third  and  fourth,  having  no  remainders,  are  leap  years  of 
three  hundred  and  sixty  six  days. 

Problem  II. 

Jo  find,  with  regard  to  any  other  years,  whether  any  given  year  be 

leap  year,  and  the  contrary. 

Rule. 

Divide  the  proposed  year  by  4,  and  if  there  be  no  remainder, 

after  the  division,  it  is  leap  year  ;  but  if  1,  2  or  3  remain,  it  is  the 

first,  second  or  third  after  leap  year. 

Examp.  1.  For  the  year  1784,         Examp.  2.  For  the  year  178G. 
4)1784(446  4)1786(446 

16  16 

18  18 

16  16 

24  26 

24  24 

~j^  "^  ^  second  after 

\  leap  year. 
Problem  III. 

To  find  the  Dominical  Letter  for  any  year,  according  to  the  Julian 

method  of  calculation. 

Rule, 

Add  to  the  year  its  fourth  part  and  4,  and  divide  that  sum  by  7  : 

if  nothing  remain,  the  Dominical  Letter  is  G  ;  but  if  there  be  any 


^ 


CHRONOLOaiCAL  PROBLEMS.  427 

remainder,  it  shews  the  letter  in  a  retrograde  order  from  G,  be- 
ginning the  reckoning  with  F ;  or,  if  it  be  subtracted  from  7,  you 
will  have  the  index  of  the  letter  from  A, accounting  as  follows: 
A  B  C  D  E  F  G 
1    2    3    4    5    C    7 
ExAMP.  For  the  year  1786. 
Given  year=1786 
Add    ^  Its  fourth  =     446 
And  4 

7)2236(319 
21 

13 

7 

66 
63 

And  7 — 3=4=t),  reckoning  from  A. 

Problem  IV. 

To  find  the  Dominical  Letter  for  any  year  according  to  the  Gregori- 
an computation. 

Rule. 
Divide  the  year  and  its  fourth  part,  less  1  (for  the  present  cen- 
tury) by  7  ;  subtract  the  remainder  after  the  division,  from  7,  and 
this  remainder  will  be  the  index  of  the  Dominical  Letter,  as  be- 
fore :  if  nothing  remain  it  is  G. 

ExAMP.  1.  For  the  year  1810.        Examp.  2.  For  the  year  1812.* 
.,,    5Givenyear=1810  1812 

-^^^    i  Its  fourth  =  452  453 


2262  2265 

Subtract         1  1 

7)2261(323  7)2264(323 

21  21 

16  16 

14  14 

21  24 

21  21 

And  7— 0=7=G.  And  7— 3=4=0. 

*  Here  it  is  to  be  observed,  that  every  leap  year  has  two  Dominical  Letters ; 
that,  found  by  this  rule,  is  the  Dominical  Letter  from  the  twenty  fifth  day  ©i 


^^ 


42S  CHRONOLOGICAL  PROBLEMS. 

Problem  V. 

To  find  the  PrimCy  or  Golden  JVumher. 

Rule. 
Add  1  to  the  given  year;  divide  the  sum  by  19,  and  the  remain- 
der, after  the  division,  will  be  the  Prime  ;  if  nothing  remain,  then 
19  will  be  the  Golden  Number. 
ExAMP.    F'or  the    year   1786. 
To  the  given  year  1786 
Add         I 

19)1787(94  H 

171 


77 
76 

1  Golden  Number. 
The  Golden  Number,  or  Lunar  Cycle,  is  a  period  of  19  years, 
invented  by  Meton,  an  Athenian,  and  from  him  called  the  Metonick, 
Cycle.  The  use  of  this  cycle  is  to  find  the  change  of  the  moon  ; 
because  after  19  years,  the  changes  of  the  moon  fall  on  the  same 
days  of  the  month  as  in  the  former  19  years  ;  though  not  at  the 
same  time  of  the  day,  there  being  an  anticipation  of  one  hour, 
twenty  seven  minutes,  forty  one  seconds,  and  thirty  two  thirds  ; 
Tvhich,  in  312  years,  amount  to  a  whole  day.  Hence,  the  Golden 
Number  will  not  show  the  true  change  of  the  moon  for  more  than 
three  hundred  and  twelve  years,  without  being  varied.  But  the 
golden  number  is  not  so  well  adapted  to  the  Gregorian,  as  the  Ju- 
lian calendar  :  The  epact  being  more  certain  in  the  new  style,  to  . 
find  which,  the  golden  number  is  of  use. 

Problem  VL 
To  find  the  Julian  Epact. 

Rule. 

First  find  the  Golden  Number,  which  multiply  by  11,  and  the 
product,  if  less  than  30,  will  be  the  number  required  ;  if  the  pro- 
duct exceed  30,  then  divide  it  by  30,  and  the  remainder  is  the 
epact. 

ExAMP.  1.  For  the  year  1786. 

February  to  the  end  of  the  year ;  and  the  next  in  tlie  order  of  the  alphabet 
serves  from  the  first  of  January  to  the  twenty  fourth  of  February. 

In  the  2d  Example,  D  is  the  Dominical  Letter  for  the  year;  but  E,  the  next 
in  the  order  of  the  alphabet,  is  th^  Dominical  Letter  for  January  and  February. 
From  this  interruption  of  the  Dominical  Letter  every  fourth  year,  it  is  twenty 
eight  years  before  the  Dominical  Letter  returns  to  the  same  order,  which,  were 
it  not  for  the  leap  years,  would  return  to  the  same  every  seven  years. 

This  Cycle  of  twenty  ei^ht  years  is  called  the  Cycle  of  the  &im. 


CHRONOLOGICAL  PROBLEMS. 


^39 


To  the  given  year  1786 
Add         1 


19)1787(94 
171 


77 
76 


Golden  Number=l  and  1x11  =  11  t^ie  Julian  Epact, 
ExAMp.  2.     For  the  vear  1791. 
1791 
1 

i9)1792(94 
171 

82 
76 

6=Golden  Number,  and  6x11=66,  therefore  30)66(2 

60 


Problem  VII. 


Epact    6 


To  find  the  Gregorian  Epact. 

Rule.  Subtract  11  from  the  JuHan  Epact:  If  the  subtraction 
cannot  be  made,  add  30  to  the  Julian  Epact ;  then  subtract,  and  the 
remainder  will  be  the  Gregorian  Epact:  if  nothing  remain,  the 
Epact  is  29. 

Or,  take  1  from  the  Golden  Number,  and  divide  the  remainder 
by  3  ;  if  1  remain,  add   10  to  the  dividend,  which  sum  will  be  the 
Epact;  if  2  remain,  add  20  to  the  dividend  ;  but  if  nothing  remain, 
the  dividend  is  the  Epact. 
ExAMP.  1.  For  the  year  1786.       Examp.  3.  For  the  year  1791. 
The  Julian  Epact  being  11      The  Julian  Epact  being  but  6 
Subtract   11  Add  to  it  30 


0 
Because  nothing  remains,  the 
Epact  is  29. 

Or, 
Examp.  2.  For  the  year  1786. 
The  Golden  number  being  I 
Take  from  it  1 

Divide  by  3)0(0 
There  being  bo  remainder, 
the  Epact  is  29,  as  before. 


Subtract  1 1 

Gregorian  Epact=25 

Or, 

Examp.  4.  For  the  year  1791. 

The  Golden  number  being  6 

Take  from  it  1 

3)5(1 
3 


430 


CHRONOLOGICAL  PROBLEM^. 


Therefore,  as  2  remains,  add  20  to  the  dividend,  and  it  gives 
the  Epact  25,  as  before. 

A  general  Rule  for  finding  the  Gregorian  Epact  forever. 

Divide  the  centuries  of  any  year  of  the  Christian  Era  by  4,  (re- 
jecting the  subsequent  numbers  ;)  multiply  the  remainder  by  17,  and 
to  this  product  add  ihe  quotient  multiplied  by  43  ;  divide  this  sura 
plus  86  by  26,  multiplying  the  Golden  Number  by  11,  from  which 
subtract  the  last  quotient,  and  rejecting  the  thirties,  the  remainder 
will  be  the  Epact. 

ExAMP.     For  the  year  1786. 

Rejecting  the  subsequent  numbers  86,  it  will  be  17. 
4)17(4 
16 

Golden  Number=  1 

1  Multiply  by  1 1 

Multiply  by  17  — 

—  11 

1 7  Subtract  the  last  quotienl=  1 1 

Add  4X43=172  — ■ 

• 00 

189  Therefore,  as  nothing  remains, 

Add    86  the  Epact  is  29,  as  before. 

23)275(11 
25 


A   TABLE    OF   THE    IvINETEEN   EPACTS    FOB,    THE   JULIAN   AND 

GREGO- 

RIAN    ACCOCTNTS,    BY 

THE    GOLDEN 

NUMBER. 

J  ulian 

Grsg. 

Julian 

Gre^. 

Julian 

Greg. 

G.  N. 

Epact. 

Epact. 

G.  N. 

Epact. 

Epact. 

G.N. 

Epact. 

Epact. 

1 

11 

29 

7 

17 

6 

13 

23 

12 

2 

22 

11 

8 

28 

17 

14 

4 

23 

3 

3  i     22   i 

9 

9 

28 

15 

15 

4 

4 

14 

3 

10 

20 

9 

16 

26 

15 

5 

25 

14 

11 

1 

20 

17 

7 

26 

6 

6 

25 

12 

12 

1 

18 
19 

18 
29 

7 
18 

Problem  VIH. 
To  calculate  the  Moon's  age  on  any  given  day. 

Rule.  To  the  given  day  of  the  month,  add  the  Epact  and  num- 
ber of  the  month  :  If  the  sum  be  less  than  30,  it  is  the  Moon's 
age,  but  if  it  exceed  30,  then  take  30  from  it,  and  the  remainder 
will  bo  the  Moon's  r.ge.  ** 

JVote.  The  numbers  to  be  added  to  the  following  months,  are 
SI  follow : 


CHRONOLOGICAL  PROBLEMS. 


431 


'January 

0 

'July        r  5- 

February 

2 

August             6 

Tu- 

March 

1 

'2' 

September,    8 

April 

October      ]    8 

May 

3 

Nove(i)ber     10 

.June 

4 

.December  LlO. 

Example.     For  January  25th,  1786. 

(Given  (lay                =25 

Add  <  Epact                       =29 

(  No.  of  the  month    =00 

64 

Subtract  30 

24=Moon's  age 

Problem  IX. 

To  find  the  times  of  the  New  and  Full  Moon,  and  the  first  and  last 

Quarters. 

Rule.  Find  the  Moon's  age  on  the  given  day,  then,  if  it  be  15, 
the  Moon  will  be  full  on  that  day,  and  by  counting  7^  days  back- 
ward and  forward  you  will  have  the  first  and  last  quarters,  and  by 
counting  backward  and  forward  15  days,  you  will  have  the  times  of 
the  last  and  next  change  ;  but  if  the  age  of  the  Moon  be  greater 
than  15,  take  15  from  it,  and  the  remainder  will  shew  how  many 
days  have  passed  since  the  last  full  moon,  and  counting  these  back- 
ward, you  will  have  the  day  the  last  full  moon  happened  on,  and  by 
knowing  that,  we  can  find  the  change,  or  either  of  the  quarters,  as 
before. 

Again,  if  the  age  of  the  moon,  on  the  assumed  day,  be  less  than 
15,  then  take  that  from  15,  and  the  remainder  will  shew  how  many 
days  are  to  run  till  the  next  full  moon,  which  you  will  have  by  ad- 
ding the  remainder  to  the  assumed  day  ;  and  proceeding  as  before, 
you  will  have  the  days  of  the  change,  and  either  quarter  as  above. 
ExAMp.  For  January  25th,  1786.    C  Assumed  day  =25 

Add  I  Epact  =29 

f  Number  of  the  month=00 


54 
Subtract  30 


Moon's  age=24 
Subtract  15 

Take  the  days  since  the  last  full  moon=  9 
From  the  assumed  day=25 

To  the  day  of  the  fall  moon=16th 
Add  15 


^i>  CHROxNOLOGICAL  PROBLEMS. 

New  Mooou  5 is: 

From  the  full  Moon  16 
Take     7^ 

First  quarter     9th 

To  the  fullMoon=16 

Add     l-l 

Last  quarter==23 

Problem  X. 

The  time  of  ihe  Moon's  coming  to  the  South,  after  the  Sun ^  being  given, 
tc  find  the  a.<re  of  the  jMoon. 

Rule.  As. 24  hours,  the  whole  difTerencc  of  time,  are  to  30,  the 
number  of  days  tVom  chatige  to  change,  so  is  the  difference  of  time, 
to  the  Moon's  age. 

Example,  I  oUserved  tlie  Moon  to  be  on  the  meridian,  or  due 
south,  at  6  o'clock  in  the  afternoon  :     What  is  the  Moon's  age  ? 

24   :   30  ::  6  :   7i  days,  Ans. 

PnonLEM  XI. 
Tojlndthe  time  of  the  Moon''s  southing. 

"Rule.  Multiply  the  Moon's  a^e,  on  the  given  day,  by  48  min- 
utes, and  divide  the  product  by  GO,  the  minutes  in  an  hour,  (or  mul- 
tiply by  4  and  divide  by  5)  and  the  quotient  will  show  how  many 
hours  and  minutes  the  moon  is  later  in  coming  on  the  meridian, 
than  the  sun,  and  counting  so  many  hours  and  minutes  forward 
from  12  o'clock,  we  have  the  time  of  the  Moon's  southing:  if  the 
liours  and  minutes,  found  as  above,  be  less  tiian  12,  then,  that  will 
be  the  lime  of  the  Moon's  southing  after  noon  ;  but,  if  greater  than 
12,  then,  take  12  from  them,  and  the  remainder  will  be  the  lime  of 
the  Moon's  southing  in  the  morning. 

KxAMP,  1.  Required  the  time  of  the  Moon's  southing  on  the 
25th  day  of  January  1786  ? 


Moon's 

age 

=24 

Or, 

h.  m. 

48 

24 

From  19  12 



4 

Take  12  00 

192 

— 



96 

5)96 

7  12 

h.  m. 

—1- 

Hence  the  Moon  60)1152(19   12      1931  =  19   12  as  before, 
souths  at  12  min-  60 

utes  past  7  in  the      , 

morn  in  2:.  552 

540 

12 


CHRONOLOGICAL  PROBLEMS.  4^3 

ExAMP.  2.     For  the  9th  of  February  1786  ? 
Moon's  age=10 
48 

h.  m. 

60)480(8  0  afternoon,  is  the  time  of  the  Moon's 
48  [southing. 

Note.  From  the  change  to  the  fall,  the  Moon  comes  to  the 
south  afternoon  ;  but  from  the  full  to  the  change,  before  noon. 

Problem  XH. 
To  find  on  what  day  of  the  week,  any  given  day  in  any  month  will  fall. 

As  one  of  the  first  seven  letters  of  the  alphabet  is  prefixed  to  eve- 
ry day  in  the  year  beginniog  with  A,  which  is  always  prefixed  to 
the  first  day  of  January  :  And  as,  in  common  years,  the  letter,  an- 
nexed to  the  first  Sunday  in  January,  shews  the  Dominical  Letter 
for  that  year  ;  but  every  leap  year  having  two  Dominical  Letters, 
the  first  of  which  serving  to  the  twenty  fourth  of  February,  and  the 
other  for  the  rest  of  the  year,  consequently,  in  any  common  year, 
the  Dominical  Letter  being  known,  the  first  of  January  may  be  ea- 
sily found,  reckoning  from  A  according  to  the  natural  order  of  the 
letters  :  and  in  any  leap  year,  the  first  of  its  two  Dominical  Letters 
will  shew  as  above,  counting  from  A  1,  B  2,  C  3,&c.  and  by  count- 
ing backward,  you  may  have  the  day  of  the  week,  on  which  the 
first  of  January  will  happen. 

Rule.  Find  the  day  of  the  week  answering  to  the  first  of  Janu- 
ary that  year,  then  add  together  the  days  contained  in  each  month 
from  the  beginning  of  the  year  to  the  proposed  day  of  the  month 
inclusively  ;  divide  this  sum  by  7,  and  if  any  thing  remain,  after  the 
division,  then  count  so  many  forward,  beginning  with  that  day  on 
which  the  first  of  January  falls,  and  you  will  have  the  day  of  the 
week,  on  which  the  proposed  day  will  fall :  but  if  nothing  remain, 
then  the  day  of  the  week,  preceding  that  day  on  which  the  first  of 
January  falls,  answers  to  the  proposed  day. 

Example. 
On  what  day  of  the  week  will  the  6th  day  of  May  1786  fall? 

Jan.        31 

By  the  preceding  observations,  and  by     Feb.       28 

Prob.  4th,  the  first  of  January  is  found  to     March   31 

fall  on  Sunday,  April      30 


Now,  counting  forward  six  days  from 
Sunday,  the  first  of  January  (inclusively) 
the  5th  of  May  falls  on  Friday. 


May         6th 

7)125(17 
7 


6/) 
49 

6  from  Jan  1 


4U- 


CHRONOLOGICAL  PROBLEMS. 


Problem  XIIL 

7h  find  the  Cycle  of  the  Sun. 

Rule.  Add  26*  to  the  given  year;  divide  the  sum  by  28,  and 
the  remainder,  after  division,  is  the  Cycle  required ;  but  if  nolh» 
ing  remain,  the  Cycle  is  28. 

Example. 
For  the  year  1807  ? 
To  1807 
Add     25 


18)1832(65 
168 

152 
140 


The  use  of  this  Cycle  is  to  find 
the  Dominical  Letter  by  the  fol- 
lowing Table. 


12=CycIe  required. 


A   TABLE    OP   THE   DOMINICAL    LETTERS    FOR    THE  NEW 
ACCORDING    TO    THE    CYCLE    OF    THE    SUN. 

STYLE, 

Cycle.    Letter.  1 

Cycle. 

Letter. 

B 
AG 

Cycle. 

Letter. 

Cycle. 

Letter. 

1 

^__ 

3 
4 
5 

6 

7 

D  C 

8 
9 

15 

"16 

17 

G 

F 

ED 

22 
23 

24 
25 

E 

D 

C 

B  A 

G 
FE 

10 
11 
12 
13 

F 

E 

D 

C  B 

18 
19 
20 
21 

C 

B 

A 

GF 

26 

27 
28 

G 
F 
E 

D 
C 

14 

A 

Problem  XIV. 

To  find  the  year  of  the  Dionysian  Period, 
Rule.     Add  to  the  given  year  457  ;  divide  the  sum  by  632,  anti 
the  remainder  will  be  the  number  required. 

Example. 
Required  the  year  of  the  Dionysian  Period  for  the  year  1786  ? 
To  1786 
Add    457 


532)2243(4 
2128 


115=Dionysian  Period. 


^'-  From  the  commeacement  of  this  century,  94-lG=r;25  must  be  added  to  the 
given  year.  The  leap  year  having  been  omitted  in  the  year  1800,  makes  it  ne- 
cessary to  add  25  to  the  date  of  the  year,  and  then  dividing  by  28,  it  will  give 
the  Cycle  right  during  the  present  century.  And  this  is  a  general  rule  to  be  ob- 
aerved,  that  w^hen  a  leap  year  has  been  abatpd,  add  16  to  the  number  whieh  wa^ 


CHRONOLOGICAL  PROBLEMS.  435 

Problem  XV. 

To  find  the  year  of  Indiction. 

Rule,  Add  3  to  the  given  year  ;  divide  the  sum  by  15,  and  the 
remainder,  after  division,  will  be  the  Indiction  ;  if  nothing  remain^ 
it  will  be  16. 

Example. 
Required  the  year  of  Indiction  for  1786  ? 
To   1786 
Add       3 


15)1789(11P 
15 

28 
15 

139 
135 

4=Indiction. 

Problem  XVI. 

To  find  the  Julian  Period. 
Rule.     Add  4713  to  the  given  year,  and  the  sum  will  be  the  Jul- 
ian Period. 

Example. 
What  year  of  the  Julian  Period  will  answer  to  the  year  1786  ? 
To     1786 
Add   4713 

6499  Ans. 

Problem  XVII. 

To  find  the  Cycle  of  the  Sun^  Golderi  Number^  and  Indictionffor  any 
current  year. 
Rule.     To  the  current  year  add  4729  ;*  divide  the  sum  by  28, 
19  and  15,  respectively,  and  the  several  remainders  will  be  the, 
numbers  required  ;  when  nothing  remains,  the  divisor  is  the  num- 
ber required. 

Example. 
What  are  the  Cycle  of  the  Sun,  Golden  Number,  and  Indiction, 
for  the  year  1807? 

before  added  to  the  year,  rejecting  28  when  it  exceeds  it,  and  this  number  be- 
ing added  to  the  year,  and  the  sum  divided  by  28,  the  remainder  after  divis- 
ion, will  be  the  Cycle  for  finding  the  Dominical  Letter.  Thus  in  the  nineteenth 
century,  it  will  be9-|-l6=25,  and  in  the  twentieth  c'entury  25-|-l6 — 28—1?; 
which  number  will  serve  two  centuries,  for  the  year  2000  is  a  leap  year. 

*  For  any  year  in  the  liiueteeoth  century  add  47 13-^- 16-— 4729, 


CHRONOLOGICAL  PROBLEMS. 


1807 
4729 

19)6536(344 
57 

15)6636(:43 
60 

28)6536(232 
56 

83 
76 

63 
45 

93 

76 

86 

84 

76 

75 

96 

0 

Jndiction=ll 

84 

Golden 

Nuinber=19 

32  Cycle  of  the  Sun. 

Problem  XVIIL 

To  find  the  time  of  High  Water, 

Rule.  Find  the  Moon's  southing,  to  which  add  the  point  of  the 
compass  making  full  sea,  on  the  full  and  change  days,  for  the  place 
proposed,  and  the  sum  will  be  the  time  required. 

Example. 
I  demand  the  time  of  high  water  at  Boston,  January  25lh,  1786, 
admitting  the  tide  to  flow  and  ebb  N.  W.  and  S.  E.  on  the  days  of 
change  ?nd  full  ? 

We  have  before  found  the  Moon^s  southing  to  be  7h.  12m.  in  the 
morning. 

h.  m. 
Therefore  to  7  12 

Add  4     0=:the  point  of  the  compass,  and  it 

Gives  11   12  in  the  morning,  for  the  time  of  high  water. 
Problem  XIX. 
To  find  on  -what  day  Easter  will  happen. 

It  was  ordered  by  the  Nicene  Council,  that  Easter  Sunday  should 
he.  kept  on  the  tirst  Sunday  after  the  first  full  moon,  which  happen- 
ed upon  or  after  the  twenty  first  day  of  March,  the  day  on  which 
they  thought  the  Vernal  Equinox  happened.  Though  this  was  a 
miJitake,  for  the  Vernal  Equinox,  that  year,  fell  on  the  twentieth  of 
March.  But  yet,  the  full  moon,  which  fell  on,  or  next  after  the 
twenty  first  of  March,  they  called  the  Paschal  full  moon.  And  by 
the  introduction  of  the  Gregorian,  or  New  Style,  the  Equinox  will 
BOW  always  happen  on  the  twentieth  or  twenty  first  of  March. 
And  the  feast  of  Easter  is  now  to  be  kept  on  the  next  Sunday  after 
the  Paschiil  full  moon,  or  the  full  moon  which  happens  after  (he 
twenty  first  of  March  ;  but,  if  the  full  moon  happens  on  a  Sunday, 
Easter  day  is  to  be  the  next  Sunday  alter. 

Rule  Find  the  age  of  the  moon  on  the  21st  of  March,  in  the 
given  year,  and  if  it  be  14,  then  find  the  day  of  the  week  answer- 
ing to  it,  and  the  Sunday  following  is  Easter  Sunday  ;  but  if  the 


CiyiONOLOGICAL  PROBLEMS.  337 

moon's  age  on  the  21st  day  of  March  he  not  14,  then  reckon  for- 
ward to  the  day  on  which  the  moon's  age  is  14,  and  find  the  day  of 
the  week  answering  to  that  day  ;  the  Sunday  following  will  be  the 
day  required. 

N.  B.     On  leap  year  take  the  20th  of  March. 

ExAMP.     When  does  Easter  happen  in  the  year  1786  ? 

21  of  March  •  Jan.       31 

29  Epact.  Feb.      28 

1  No.  of  the  month.  March    31 

April      13lh 

51  

Subt.'30  7)103(14 

7 
21  Moon's  age.  — 

AA\  Q-^S  ^^'  ^^  ^^y^  ^°  ^^®  Moon's  33 

Add  ^d  ^  ^^ji^^  14  days  old.  28 

44  6     There - 

Take  31=days  in  March,     fore  the  first  of  January  being  Sun- 

—  day,  reckon  forward  5  days,  includ- 

13th  of  April,  the        ing  Sunday,  and   you  will  find  the 

day   of  the    full  moon,  or        13lh    of  April   falls   on  Thursday, 

Easter  limit.  consequently  the  next  Sunday  is  the 

16th,  which  is  Easter  Sunday. 
Easter  may  be  found,  for  any  future  time,  by  the  following  Ta- 
ble which  is  calculated  from  1753,  the  time  of  the  commencement 
of  the  New  Style  in  America,  and  which  shews,  by  the  Goldeft 
Number^  the  days  of  the  Paschal  full  moons ;  by  which,  and  the 
Dominical  Letter,  the  day  on  which  Easter  will  fall,  may  be  found, 

Tke  Use  of  the  Table. 

First,  find  the  Golden  Number  as  before  taught,  which  seek  in 
the  column  of  Golden  Numbers  under  the  time  in  which  the  given 
year  is  included  ;  right  against  the  Golden  Number  of  the  year,  in 
the  last  column  but  one,  you  have  the  day  of  the  month  on  which 
the  Paschal  full  moon  happens,  which  is  the  limit  of  Easter  ;  from 
thence  run  your  eye  down  among  the  Dominical  Letters,  till  you 
come  to  the  Letter  of  the  given  year,  and  against  it  you  have  the 
day  of  the  month,  on  which  Easter  falls  that  year. 

Example.  To  know  when  Easter  falls  in  1786. 

The  Golden  Number  for  the  year  being  one,  and  the  Dominical 
Letter  A  ;  therefore  seek  in  the  first  column  (the  given  year  be- 
ing  included  between  the  years  1753  and  1899)  for  the  Golden 
Number:  then  cast  your  eye  along  to  the  last  column  but  one,  un- 
der the  title  Paschal  full  i^,  and  you  will  find  the  thirteenth  of 
April  to  be  the  day  of  the  full  moon  ;  against  which,  in  the  last 
column,  stands  E,  which  shews  it  to  be  Thursday,  therefore  the 
next  Sunday  following  is  Easter  Sunday,  which,  by  going  down  the 
column  of  Letters  to  the  next  A,  you  will  find  to  be  the  sixteenth 
of  April. 


433 


TABLE  OF  GOLDEN  NUMBERS. 


GOLDEN 

JVUMBERS 

FRO  31 

1753  TO  1 

a99 

AND    SO    ON    TO    4199.  t 

INCLUSIVELY. 

1 

i 

O 

^ 

1 

-  -f- 

s 

o 

8  i 

05 

g 

OS 

O 

oo    oo 

g  § 

o   o 

03 

c 
c 

OS 
CO 

o 

O 

►ft' 

i 

—  rs 

O 

o 

3 

o     o 

r' 

O 

c 

r- 

sir 

o 

o 

o 

o" 

o 

o 

o 

ml 

r- 

CO 

CO 
CO 

CO 

CO 
CO 

CO 

<0 

CC' 

CO 

;o 

CO 

CO 
CO 

CO 
CO 

CO 

CO 

CO 

CO 
CO 

03 

c:: 

CO 
CO 

02 

-1 

CO 

CO 

4^ 

S 

CD 

CO 
CO 

-J 

Ti 

— 

~G 

n 

6 

Ti 



~9 

n 

1 

Ti 

~T 

T2 

— 

~4 

21 

(T 

3 

14 

— 

6 

— 

6 

17 



9 

— 

9 

— 

1 

12 

— 

•g22 

D 

— 

3 

11 

— 

14 

— 

6 

17 



9 

— 

9 

— 

1 

12 

-23 

E 

11 

— 

3 

14 

3 

14 

_- 

6 

17 

— 

9 

— 

9 

— 

1 

S24 

F 

■— 

11 

— 

J 

— 

14 

t) 

\\ 

■— 

17 

— 

9 

25 

G 

19 

— 

TT 

— 

11 

~ 

13 

T4 

~ 

6 

17 

6 

17 

Zl" 

"9 

26 

F 

8 

19 

— 

11 

— 

11 

.... 

3 

14 

— 

6 

6 

17 

— 

27 

B 

— 

8 

19 

-- 

19 



11 



3 

14 

14 

— 

6 

17 

28 

C 

16 

— 

8 

19 

8 

19 

— 

11 



3 

14i   3 

14 

— 

6 

29 

D 

5 

16 

8 

— 

8 

19 

11 

— 

J 

— 

3 

14 

— 

30 

E 

13 

5 

16 
~5 

16 

16 
~5 

T6 

8 

19 
~8 

T9 

11 

TT 

11 

TT 

3 

14 
~3 

31 

F 

1 

cT 

2 

13 

— 

5 

— 

5 

16 



8 

19 

— 

19 

— 

11 

— 

.-r    2 

A 

— 

2 

13 



13 

— 

5 

16 



8 

19 

8 

19 

11 

5-  3 

B 

10 

— 

2 

13 

2 

13 

— 

5 

16 

— 

8 

8 

19 

<    4 

C 

Ta 

10 

-- 

To 

2 

To 

2 

13 

2 

13 

5 

18 
~5 

To 

16 
~5 

Te 

8 

19 

8 

5 

D 

6 

E 

1 

18 

— 

10 

— 

10 

— 

2 

13 

— 

6 

5 

16 

7 

F 

— 

7 

18 

— 

18 

— 

10 



2 

13 

— 

13 

— 

5 

16 

8 

G 

15 

— 

7 

18 

7 

18 

— 

10 



2 

13 

2 

13 

6 

9 

A 

4 

15 

— 

7 

— 

7 

18 

— 

10 

— 

2 

2 

13 

10 

B 

— 

~4 

15 

— 

15 

IZi 

~7 

T8 

"~ 

To 

— 

To 

— 

1 

T3 

11 

c" 

12 

— 

4 

15 

4 

15 

— 

7 

18 

— 

10 

— 

10 

— 

2 

12 

D 

1 

12 

._ 

4 

— 

4 

15 



7 

18 

— 

18 

— 

10 

13 

E 

— 

1 

12 

— 

12 

■ 

4 

15 

— 

7 

18 

7 

18 

— 

10 

14 

F 

9 

~9 

1 

12 
~T 

1 

12 
1 

Ti 

4 

15 

4 

T5 

7 

T5 

7 

18 

15 

G 

~7 

Ts 

16 

17 

17 

9 

— 

9 

— 

1 

12 

12 

4 

15 

4 

15 

15 

7 

17 

B 

6 

6 

17 

9 

17 

9 

9 

1 

1 

12 

4 

12 

4 

4 

15 

18 

C 

19 

j 

20 

E 

21 

F 

22 

G 

23    -A 

25    C 

GEOMETRICAL  PROBLEMS,  i39 


PLANE  GEOMETRY. 


DEFINITIONS. 

L  A  POIJVTin  the  Mathematicks  is  considered  only  as  a  mark, 
without  any  regard  to  dimensions. 

2.  A  Line  is  considered  as  length,  without  regard  to  breadth  or 
thickness. 

3.  A  Plane  or  Surface  has  two  dimensions,  length,  and  breadth, 
but  is  not  considered  as  having  thickness. 

4.  A  Solid  has  three  dimensions,  length,  breadth  and  thickness, 
and  is  usually  called  a  Body. 

5.  A  line  is  e'lihev  straight ^  which  is  the  nearest  distance  between 
two  Points ;  or  crooked,  called  a  Curve  Line,  whose  ends  may  be 
drawn  further  asunder. 

6.  If  two  Lines  are  at  equal  distance  from  one  another  in  every 
part,  they  are  called  parallel  Lines,  which,  if  continued  infinitely, 
will  never  meet. 

7.  If  two  lines  incline  one  towards  another,  they  will,  if  continu- 
ed, meet  in  a  point :  by  which  meeting  is  formed  an  Angle. 

8.  If  one  Line  fall  directly  upon  another,  so  that  the  Angles  on 
both  sides  are  equal,  the  Line,  so  falling,  is  called  a  perpendicitlary 
and  the  Angles  so  made,  are  called  right  Angles,  and  are  equal  to 
90  degrees,  each. 

9.  All  Angles,  except  right  Angles,  are  called  oblique  Angles? 
whether  they  are  acwie,  that  is,  less  than  a  right  Angle  ;  or  obtuse, 
Chat  is,  greater  than  a  right  Angle. 

GEOMETRICAL  PROBLEMS. 

Problem  I.     To  divide  a  Line  AB  into  tzvo  equal  parts. 

Set  one  foot  of  the  compasses  in 
the  point  A,  and,  opening  them  be- 
yond the  middle  of  the  line,  de- 
scribe arches  above  and  below  the 

line  ;  with  the  same  extent  of  the  ^^ 

compasses,  set  one  foot  in  the  point  A  ^» 

B,  and  describe  two  arches  crossing 
the  former  :  draw  a  line  from  the 
intersection  of  the  arches  above  the 
line,  to  the  intersection  below  ihe 
line,  and  it  will  divide  the  line  AB  into  two  equal  parts. 


4 

/  *♦ 


446 


GEOMETRICAL  PROBLEMS. 


Problem  IL  To  erect  a  perpendicular  on  the  point  C  in  a  given  hne. 


X 


Set  one  foot  of  the  compasses 
in  the  giv*en  point  C,  extend  the 
other  foot  to  any  distance  at  pleas- 
ure, as  to  D,  and  with  that  extent 
make  the  marks  D,  and  E.  With 
the  compasses,  one  foot  in  D,  at 
any  extent  above  half  the  distance 
of   D   and    E,   describe  an  arch  , 

above  the  line,  and  with  the  same         ^  C  K' 

extent,  and  one  foot  in  E,  describe 
an    arch   crossing    the    former ; 

draw  a  line  from  the  intersection  of  the  arches  to  the  given  point 
C,  which  will  be  perpendicular  to  the  given  line  in  the  point  G. 


Problem  III.     To  erect  a  perpendicular  upon  the  end  of  a  line. 


o  ....- 


K' 


_lB 


'K 


Set  one  foot  of  the  compasses 
in  the  given  point  B,  open  them 
to  any  cofivenient  distance,  and 
describe  the  arch  CDE  ;  set  one 
foot  in  C,  and  with  the  same  ex- 
tent, cross  the  arch  at  D  :  with 
the  same  extent  cross  the  arch 
again  from  D  to  E  ;  then  with 
one  foot  of  the  compasses  in  D, 
and  with  any  extent  above  the 

half  of  ED,  describe  an  arch  a  ;  take  the  compasses  from  D,  and, 
keeping  them  at  the  same  extent  with  one  foot  in  E,  intersect  the 
former  arch  a  in  a ;  from  thence  draw  a  line  to  the  point  B,  which 
will  be  a  perpendicular  to  AB. 

Problem  IV.     From  a  given  pointy  a,  to  let  fall  a  perpendicular  to 
a  given  line  AB. 

Set  one  foot  of  the  compasses  in 
the  point  a,  extend  the  other  so  as 
to  reach  beyond  the  line  AB,  and 
describe  an  arch  to  cut  the  line  AB 
in  C  and  D  ;  put  one  foot  of  the 
compasses  in  C,  and,  with  any  ex- 
tent above  half  CD,  describe  an 
arch  b  ;  keeping  the  compasses  at 
the  same  extent,  put  one  foot  in  D,  A 
and  intersect  the  arch  b  m  h  \ 
through    which   intersection,  and 


V 


/  K 


di 


-E O 


B 


the  point  fi,  draw  a  E,  the  perpendicular  required. 


GEOMETRICAL  PROBLEMS. 


441 


Problem  V.     To  draw  a  Line  parallel  to  a  given  Line  AB. 
Set  one  foot  of  the  com- 

E  ^'"  >".. — ^.-^  _  *'. 'F 


passes  in  any  part  of  the  line, 
as  at  c  ;  extend  the  compass- 
es at  pleasure,  unless  a  dis- 
tance be  assigned,  and  de- 
scribe an  arch  6;  with  the 
same  extent  in  some  other 


^ 


T 


B 


part  of  the  line  AB,  as  at  e,  describe  the  arch  a  ;  lay  a  ruler  to  the 
extremities  of  the  arches,  and  draw  the  line  E  F,  which  will  be 
parallel  to  the  line  A  B. 

Problem  VI.     To  make  an  Angle  equal  to  any  number  of  degrees* 

It  is  required  to  lay  off  an  acute  Angle  of  35°  on  a  given  line  AB. 

Take  60 degrees  frorp  the  line  of 
chords  in  the  compasses,  set  one 
foot  of  the  compasses  in  the  point 
A,  describe  an  arch  CD,  at  pleas- 
ure ;  then  set  one  foot  of  the  com- 
passes in  the  brass  centre,  in  the 
beginning  of  the  line  of  chords,  and 
bring  the  other  to  35  on  the  line  ; 
with  this  extent  set  one  foot  in  C, 

with  the  other  intersect  the  arch  CD,  in  a,  and  through  a  draw  the 
line  AE,  so  will  EAB  be  an  angle  of  35  degrees. 

If  the  angle  had  been  obtuse,  suppose  125°,  then  take  90°  from 
the  line  of  chords  ;  set  one  foot  in  C,  and  intersect  the  arch  in  b ; 
then  take  35°  from  the  same  line  of  chords,  and  set  them  from  b  to 
di  a  line  drawn  from  A  through  d  to  F  will  make  an  angle,  FAB, 
of  125°. 

To  measure  an  angle  by  the  line  of  chords,  is  only  to  take  the 
distance  on  the  arch  between  the  lines  AB  and  AE,  or  AB  and  AF, 
and  lay  it  on  the  line  of  chords. 

Problem  VII.  To  make  a  Triangle^  whose  sides  shall  be  equal  to  three 
§iven  lines,  provided  any  two  of  them  be  longer  than  the  third. 


Let  A,B,C,  be  the  three  given  lines ; 
draw  a  line  AB,  at  pleasure  ;  take 
the  line  C  in  the  compasses,  set  one 
foot  in  A,  and  with  the  other  make  a 
mark  at  B  ;  then  take  the  given  line 
B  in  the  compasses,  and  setting  one 
foot  in  A,  draw  the  arch  C  ;  then 
take  the  line  A  in  the  compasses,  and 
intersect  the  arch  C  in  C ;  lastly, 
draw  the  lines  AC  and  BC,  and  the 
friangle  will  be  completed. 


H  r^ 


442 


GEOMETRICAL  PROBLEMS. 


to  am; 


Proelem  VIIL     To  make  a  Square^  having  equal  sides,  equal 

given  line. 

Let  A  be  the  given   line  ;  draw  a  line        a 
AB  equal  to  the  given  line  ;  from  P  raise       — — — 
a  perpendicular  to  C  pqual  to  AB,  with  _  ■' 
the  same  extent,  set  one  foot  inCandde 
scribe  the  arch  D  ;  also  with  the  same  ex- 
tent, set  one  foot  in  A  and  intersect  the 
arch  D  ;  lastly,  draw  the   lines  AD  and 
CD,  and  the  square  will  be  completed. 

In  like  manner  may  a  Parallelogram  be 
constructed,  only  attending  to  the  differ- 
ence between  the  length  and  breadth.      A 

Problem  IX.  To  describe  a  Circle,  which  shall  pass  through  any 
three  given  Points,  which  are  not  in  a  straight  line. 
Let  the  three  given  points  be  A,B,C,  through  which  the  circle  is 
to  pass.  Join  the  points  AB  and  BC  with  right  lines,  and  bisect 
these  lines;  the  point  D,  where  the  bisecting  lines  cross  each  oth- 
er, will  be  the  centre  of  the  circle  required.  Therefore,  place 
one  point  of  the  compasses  in  D,  extending  the  other  to  cither  of 
the  given  points,  and  the  circle,  described  by  that  radius,  will  pas? 
through  all  the  points. 


Hence,  it  will  be  ea«>y  to  find  the  cen- 
tre of  any  given  circle  ;  for,  if  any  three 
points  are  taken  in  the  circumference  of 
the  given  circle,  the  centre  will  be  rea- 
dily tbnnd  as  above.  'J'he  same  may  al- 
so be  observed,  when  only  a  part  of  the 
circumference  is  given. 


Problem  X.     To  describe  an  Ellipsis  or  Oval  mechanically. 

Dravv  two  parallel  lines,  as  L  and  M,  at  a  moderate  distance,  by 
Prob.  5  ;  then  draw  two  others  at  the  same  distance,  across  the  for- 
mer, as  N  and  O  ;  by  the  crossing  of  these  lines  will  be  made  a  fig- 
ure ABCD,  of  four  sides  ;  extend  the  compasses  at  pleasure,  and  set- 
ting one  foot  in  D,  describe  the  arch  cde  ; 
with  the  same  extent,  set  one  foot  in  B, 
and  describe  the  arch  fgh  ;  then  set  one 
foot  in  C,  and  contract  them  so  as  to  reach 
the  point  c,  and  describe  the  arch  Im  ; 
with  the  same  extent,  and  one  foot  in  A, 
describe  the  arch  ik,  and  the  oval  will  be 
completed.  In  the  same  manner,  with 
a  greater  or  less  extent  of  the  compass- 
es, may  a  greater  or  less  oval  be  made  by  the  same  four  sided  fig- 
ure A  BCD. 


MENSURATION  OF  SUPERFICIES,  Lc,  443 

MENSURATION 

OF  SUPERFICIES  AND  SOLIDS. 
Section  I.    Of  Superficies. 

SUPERFICIES,  or  surfaces,  are  measured  by  the  superficial 
inch,  foot,  yard,  &c.  according  to  the  measures  peculiar  to  differ- 
ent artists. 

The  superficial  inch,  foot,  &c.  is  one  inch,  foot,  &c.  in  length  and 
breadth;  and,  because  12  inches  make  one  foot  of  Long  Measure, 
therefore  12x12=144  inches  make  1  superficial  foot,  3x3=9  feet, 
a  yard,  &c. 

The  superficial  content  of  every  surface  is  found  by  the  proper 
rule  of  its  figure,  whether  square,  triangle,  polygon,  or  circle. 
Article  1.   To  measure  a  Square y  having  equal  sides. 

Rule. 
Multiply  the  side  of  the  square  into  itself,  and  the  product  will 
be  the  area  or  superficial  content,  of  the  same  name  with  the  de- 
nomination taken,  either  in  inches,  feet,  or  yards,  respectively. 

Let  ABCD  represent  a  square,  whose  side  is  12   a 
inches  or  12  feet.     Multiply  the  side  12  by  itself,       ' 
thus,  12  inches.  12  feet. 

12  inches.  12  feet. 

Area=144  inches.  144  feet. 

Bij  the  Sliding  Rule. 

Set  1  to  the  length  on  B,  then,  find  the  breadth  on  A,  and  oppo 
site  to  this  on  B,  you  will  have  the  content. 

By  Guntcr^s  Scale. 

Extend  the  dividers  from  1,  on  the  line  of  numbers,  to  the  length  ; 
that  distance,  laid  the  same  way  from  the  breadth,  will  point  out 
the  answer. 

Art.  2.  To  measure  a  Parallelogram  or  long  Square. 

Rule. 
Multiply  the  length  by  the  breadth,  and  the  product  will  be  the 
area,  or  superficial  content* 

Let  ABCD  represent  a  parallelogram,  whose  A  B 

length  is  5  feet,  and  breadth,  4  feet.     Multiply 
5  by  4.         Length    5 
Breadth  4 

_  D 

Area  20  Ans. 

*  If  the  parallelogram  be  divided  into  squares  by  drawings  lines  as  in  the 
figure,  it  is  obvious  on  inspection,  that  the  number  of  squares  must  always  be 
equal  to  the  product  of  the  length  and  breadth.  The  same  may  be  shown  ou 
the  square  also.  The  area  of  a  Rhombus  or  Rhomboides  is  equal  to  that  of  a 
T)4iraUeIogram  of  the  same  base  and  altitude. 


444  MENSURATION  OF  SUPERFICIES 

The  content  of  this  figure  is  found  on  the  sliding  rule  and  scale, 
as  the  former. 

Art.  3.   When  the  breadth  of  a  Superficies  is  given^  to  find  how  much 
in  length  will  make  a  square  foot^  yardy  ^c. 

Rule. 
As  the  breadth  is  to  a  foot,  yard,  &c.  so  is  a  foot,  yard,  &c.  to 
the  length  required  to  make  a  foot,  yard,  &,c.     Or  divide  144  by 
the  breadth,  and  the  quotient  will  be  the  length  required. 

How  much,  in  length,  of  a  board  21  feet  wide,  will  make  a  square 
foot? 

In.  br.  In.  leng.  In.  br.  In.  leng. 
As  30  :    12  ::    12  :  48 
12 

30)144(4-8  inches,  length  required. 
120 

240 

240  In. 

Breadth=30)144(4-8  inches,  Ans. 

Art.  4.  To  measure  a  Rhombus. 

JJefmiiion,  A  rhombus  is  a  figure  with  four  equal  sides,  in  the 
form  of  a  diamond  on  cards,  having  two  angles  greater  and  two  less 
than  the  angles  of  a  square  :  the  former  are  called  obtuse  angles, 
and  the  latter,  acutey  or  sharp,  angles. 

Rule. 

Multiply  the  side  by  the  length  of  a  perpendicular,  let  fall  from 
one  of  the  obtuse  angles  to  the  side  opposite  such  angle. 

Let  ABCD  represent  a  rhombus, 
each  of  whose  sides  is   16  feet:  AAr  ^^ 

perpendicular  let  fall  from  the  obtuse     \ 
angle,  at  B,  on  the  side  DC,  will  in-       \ 
tersect  it  in  the  point  E,  so  will  BE         \ 
be  12  feet ;  and  this  being  multiplied  \ 

into  the  given  side,  the  product  will  \ 

be  the  area  of  the  rhombus.  T>V 

yV     r^ ^V 

Side=16  By  the  Sliding  Rule. 

Fer.=12  Set  1   on  A  to  the   length  on  B;  find  the 

perpendicular  height  on  A,  against  which  on 

192  area.  B  is  the  content. 

By  Gunter. 

The  extent  from  1  to  the  perpendicular  height  will  reach  from 
the  length  to  the  content. 


AND  SOLIDS. 


44c 


Art.  5.  To  find  the  Area  of  a  Rhomhoides. 

Defirtntion.  A  rhomboides  is  a  figure,  whose  opposite  sides  and 
opposite  angles  are  equal. 

Rule. 
Multiply  one  of  the  longest  sides  by  the  perpendicular  let  fal: 
from  one  of  the  obtuse  angles  on  one  of  the  longest  sides. 

Let  ABCD  represent  a  rhomboides ; 
the  longest  sides  AB  and  CD  being  /;  /B 

16-5  feet,  and  the  perpendicular  AE, 
■"^•7  feet.      Side=:16-5  jO       __ 

Perp.    9-7  ^ 

The  content  is  found  on  the 
sliding  rule,  and  scale,  as  in 
the  last  figurp. 
Ans.  16005  feet. 

Art.  6.   To  measure  a  Triangle.* 
Rule. 

If  it  be  a  right  angled  triangle,  multiply  the  base  by  half  the 
perpendicular,  or  half  the  base  by  the  perpendicular,  and  the  pro- 
duct wiH  be  the  area:  but  if  it  be  an  oblique  angled  triangle, 
(whether  obtuse,  or  acute,)  multiply  half  the  base  by  the  length 
of  the  perpendicular  let  fall  on  the  base  from  the  angle  opposite  to 
it,  and  the  product  will  be  the  area.  The  longest  side  of  a  irian- 
gle  is  usually  called  the  base,  except  in  a  right  angled  triangle, 
where  the  longest  of  the  two  legs,  which  include  the  right  angle, 
is  called  the  base. 

In  the  right  angled  triangle  ABC  right 
angled  at  C  ;  the  base  AC  is  18-8  feet,  and 
the  perpendicular  BC=12-6. 

Base     =18-8         Or,  Perp.=12-6 


jPerp.=  6-3 


Base 


664 
1128 


9-4 

504 
1134 


11844  area.  118-44  area 

The  oblique  angled  triangle  ABC 
being  given,  let  fall  a  perpendicular 
from  the  angle  at  B  on  the  base  AC, 
and  that  perpendicular  is  the  height 
of  the  triangle.  The  base  AC  being  i 
15-6,  and  the  perpendicular  BD=9,  ■ 
to  find  the  area. 

*  A  triangle  is  half  a  parallelogram  of  the  same  base  and  altitude  ;  hence  th« 
rule.  In  a  right  angled  triangle,  the  longest  side  is  called  the  hypotenuse ;  the 
next  longest,  the  oase ;  and  the  shortest  side,  the  perpendiculjxr. 


446  MENSURATION  OF  SUPERFiCIEt 

78=half  the  base. 
9=height  of  the  angle. 


70-2=area. 

By  the  Sliding  Rule. 

Set  1  on  A  to  the  length  of  the  base  on  B,  and  opposite  to  half 
the  length  of  the  perpendicular,  on  A,  you  will  have  the  content 
on  B. 

By  Gunter. 

The  extent  from  1  to  half  the  length  of  the  perpendicular  will 
reach  from  the  length  of  the  base  to  the  content. 

In  this  place  it  may  be  proper  to  instruct  the  learner  in  one  of 
the  properties  of  a  right  angled  triangle  :  viz  That  the  j'quare  of 
the  longest  side  of  a  right  angled  triangle,  usually  called  the  hy- 
potenuse, is  equal  to  the  sum  of  the  squares  of  the  two  other  sides, 
usually  called  the  legs  ;  which  is  of  great  use,  for  by  this  mean, 
any  two  sides  of  a  right  angled  triangle  being  given,  the  other  may 
be  found  by  common  Arithmetick.  Thus,  in  the  right  angled  tri- 
angle ABC,  the  base  AC  and  perpendicular  BC  being  given,  the 
hypotenuse  AB  may  be  found  by  extracting  the  square  root  of  the 
sum  of  the  squares  of  the  ba^e  and  perpendicular. 

Base   18  8  Perp.   12  6  353-44=square  of  the  base. 

18-8  12  6  158-76=squareoftheperp. 


1504 
1504 
188 

756 
252 
126 

61 2-20(22'63  hypotenuse 
4 

353-44 

158-76 

42)112 
84 

446)2820 

2676 

• 

4523)14400 
13569 

831 

And,  if  the  hypotenuse  and  one  of  the  legs  be  given,  the  other 
may  be  found  by  subtracting  the  square  of  the  given  leg  from  the 
square  of  tiie  hypotenuse. 

There  are  some  numbers,  the  sum  of  whose  squares  make  a  per- 
fect square,  of  which  sort  are  3  and  4,  whose  squares,  being  added 
together,  make  25,  which  is  Ih^  square  of  5 ;  therefore,  if  the  base 
of  a  triangle  be  4,  and  the  perpendicular  3,  the  hypotenuse  will  be 
5  ;  and  if  any  of  these  numbers  be  multiplied  by  any  other  number, 
those  products  will  be  the  j^ides  of  right  angled  triangles,  as  6,  8,  10, 
and  15,  20,  S5,  &c  Thus  artiticers,  when  they  set  off  the  corner 
of  a  building,  usually  measure  6  feet  on  one  side,  and  8  feet  on  the 
other,  then  laying  a  10  feet  pole  across,  it  makes  the  corner  a  true 
right  angle. 


AND  SOLIDS. 


44' 


Art.  7     I'here  is  another  method  of  finding  the  area  of  triangles,  the 
three  sides  being  given. 

Rule.  Add  the  three  sides  together,  then  take  the  half  of  that 
sum,  and  out  of  it  subtract  each  side  severally  ;  and  muhiply  ihe 
half  of  the  sum  and  these  remainders  continually,  and  the  square 
root  of  this  product  will  be  the  area  of  the  triangle. 

In  the  oblique  triangle  ABC,  the  base  AC  is  given  loG,  the  side 
AB  is  10  4,  and  the  side  BC  is  9-2,  to  find  the  area. 

15  6  17-6  17-6  17-6  o     .. 

10-4  — 16-6         —10  4         —  92 


8-4 


35  2  sum 


^C 


17-6=half  the  sum. 
.  17-6 

2 

35-2 

7*2 

704 
2464 

253-44 
84 

101376 
202752 

2128-896 


2128-8960(46'139=area. 
16 


86)528 
516 

921)1289 
921 

9223)36860 
27669 


92269)919100 
830421 


88679 


Art.  8.     To  measure  a  Trapezium. 

Definition.  A  trapezium  is  an  irregular  figure  of  four  unequatl 
sides,  and   uwequal  angles. 

Rule.  Draw  a  diagonal  line  from  one  of  the  angles  to  the  oppo- 
site angle,  as  AC,  and  then  will  the  trapezium  be  divided  into  two 
triangles,  of  which  the  diagonal  is  the  common  base  :  then,  letting 
fall  perpendiculars  from  the  other  opposite  angles  on  the  diagonal^ 
add  those  perpendicuUrs  tof^ether,  and  multiply  half  that  sum  into 
the  diagonal,  or  half  of  the  diagonal  into  the  sum  of  the  perpendic- 
ulars, and  that  product  will  be  the  area  of  the  trapezium. 

In  the  trapezium  ABCD,  the 
cliagonal  AC  is  24,  the  perpen- 
dicular DE  0,  and  the  perpendi- 
cular EF  10.  The  sum  of  the 
perpendiculars  is  16,  whose  half 
is  8,  which  being  multiplied  in* 
to  24,  will  give  the  area. 


^41 


MENSURATION  OF  SUPERFICIES 


24 
8 


192=area. 

By  the  sliding  Rule. 

Set  1  on  A  to  I  the  sum  of  the  perpendiculars  on  B,  and  opposite 
the  length  of  the  diagonal  on  A,  you  will  have  the  area  on  B. 

By  Gunter. 

The  extent  from  1  to  ^\he  sum  of  the  perpendiculars  will  reach 
from  the  length  of  the  diagonal  to  the  area. 

Art.  9.     To  measure  any  irregular  figure. 

Rule.  Divide  the  figure  into  triangles,  by  drawing  diagonals 
from  one  angle  to  another;  then  measure  all  the  triangles  by  ei- 
ther of  the  rules,  already  taught,  at  Article  6  or  7,  and  th.e  sum  of 
the  several  areas  of  all  the  triangles  will  be  the  area  of  the  given 
fiorure. 


The  irregular  figure  ABCDEF 
being  given,  divide  it  into  triangles 
by  the  diagonals  FB,  EB,  and  DB  : 
then  may  the  triangles  be  meas- 
ured by  letting  fall  perpendiculars 
on  their  respective  bases,  as  Ba, 
B6,  Dc,  ¥d,  and  multiplying  those 
perpendiculars  by  half  their  res- 
pective bases. 


In  the  triangle  AFB  the  base  FA  is  loo,  and  the  perpendicular 
Ba49  ;  in  the  triangle  FBE  the  base  BE  is  9:2,  and  the  perpendi- 
cular Yd  52 ;  Hi  the  triangle  EBD,  the  base  BE  is  the  same  as  be- 
fore,  and  the  perpendicular  Y)c  44;  and  in  the  triangle  DCB,  the 
base  DC  is  80,  and  the  perpendicular  B6  38  ;  by  which  the  area 
of  each  may  be  found  by  Art.  6,  as  follows. 


60=half  AF. 
49=perp.  aB, 


46=ha!f  BE. 
52=perp.  Yd, 


2450=area  of  AFB. 

46=halfBE. 
44=perp.  Dc. 

184 

184 


9'. 
230 


2392=area  of  FBE. 

38=perp.  B&. 
40=halfDC. 


2450 
2024 
2392 
1520 

838G=area  of  the 
figure  ABCDEF. 


2024=area  of  EBD.    1520=area  of  DCB. 


AND  SOLIDS. 


44& 


In  dividing  any  irregular  figure  into  triangles,  the  triangles  will 
be  less,  by  two,  and  the  diagonals  less,  by  three,  than  the  numbey 
of  the  sides  of  the  figure. 

If  there  be  a  long,  irregular 
figure  like  the  following,  the 
mean  breadth  may  be  found 
very  nearly,  by  measuring  the 
breadth  at  certain  equal  distan- 
ces along  AB,  and  dividing  the 
sum  of  the  breadths  by  their 
number. 

Let  the  length,  AB,  be  16  rods,  the  1st  breadth  AC  3  9  rods,  the 

2d  4  rods,  the  3d  3  93  rods,  the  4th  4-3  rods,  the  5th  4-25  rods, 

the  6th  4-5  rods,  the  7th  48  rods,  and  the  8th  4  9  rods  ;  what  is 

the  area  ? 

3^9-f4--f.3-96+4-3+4-254-4-6+4-8-{-4  9       34-6 
_ ~  __    jjjg   mean 

34-6 
breadth.     Then —^Xl6=69-2  rods,  Ans. 


Art.  10.     7o  measure  a  Trapezoid. 

Definition.  A  trapezoid  is  the  segment  of  a  triangle,  cut  by  a 
line  parallel  to  the  base. 

Rule.  Add  the  parallel  sides  together,  and  multiply  half  that 
sum  by  the  perpendicular  breadth. 


In  the  trapezoid 
ABCD,thesideAD 
18  24,  the  side  BC  is 
16,  and  the  perpen- 
dicular breadth  Ba 
is   10,  to    find    the 


24= AD 
16=BC 

40=sum. 


-Ad 


20= 

10: 


■  1 
"2 

Ba. 


sum. 


area  by  adding  the 

sides  BC    and    AD 

and  multiplying  half  200=area. 

their  sum  by  the  perpendicular  breadth  Ba. 

By  the  Sliding  Rule. 

Set  1  on  A  to  the  equated  length  on  B,  and  against  the  breadth 
on  A  you  will  have  the  area  on  B. 

By  Gunter. 

The  extent  from  1  to  the  breadth  will  reach  from  the  equated 
length  to  the  area. 

Art.  11.     To  measure  any  regular  Polygon. 

Definition.  A  regular  polygon  is  a  figure  whose  sides  and  an- 
gles are  all  equal ;  they  are  usually  denominated  from  the  number 
of  their  sides. 

1  3 


*■ 


450 


MENSURATION  OF  SUPERFICIES 


Thus,  A  figure  having 


equal  sides  and 
angles  is  a 


'Trigon. 
Tetragon. 
Pentagon. 
Hexagon. 
Heptagon* 
Ocfagon. 
Enneagon. 
Decagon. 
Endecagon. 
Dodecagon. 


Rule.  Blultiply  the  length  of  one  of  the  sides  by  the  number 
of  sides  ;  then,  this  product  by  the  half  of  a  perpendicular  let  fall 
from  the  centre  of  the  figure  to  the  middle  of  one  of  the  sides, 
and  the  product  will  be  the  area  of  the  polygon. 

In  the  pentagon  ABODE,  each  side  is  95, 
and  the  perpendicular  FG  65  36,  to  find  th,e 
area.  95=lei>gth  of  a  side. 

6=number  of  sides.  '^* 

475=:sum  of  the  sides. 
32-68— i  of  the  perpendicular. 


15623-00=area  of  the  pentagon. 

By  ike  Sliding  Rule. 
Set  1  on  A  to  J  the  perpendicular  on  B,  and  against  the  sum  of 
the  sides  on  A  you  will  have  the  area  on  B. 

By  Gunter. 

The  extent  from  I  to  half  the  length  of  the  perpendicular,  will 
reach  from  the  sum  of  the  sides  to  the  content. 

But  for  the  more  ready  measuring  regular  polygons,  the  follow- 
ing Table,  containing  multipliers  for  all  regular  figures  from  the 
triangle  to  the  dodecagon,  will  be  of  use  to  the  learner. 


Number 
•of  sides. 

Names. 

Multipliers. 

Number 
of  sides. 

Name?. 

Multipliers. 

3 
4 
5 
6 

7 

Trigon. 

Tetragon. 

Pentagon. 

Hexagon. 

Heptagon 

•433oiar 

1- 

1-720477 
2-589076 
3-633959 

8 

9 

10 

11 

12 

Octagon. 

Enneagon. 

Decagon. 

Endecagon. 

Dodecagon. 

4-828427 
6-181827 
7-694209 
9-361 
11196 

If  the  square  of  the  sidr  of  a  polygon  be  multiplied  by  the  mul- 
tipli«r  of  the  like  figure,  the  product  will  be  the  area  of  the  figure 
isought. 


AND  SOLIDS.  45i 

To  measure  a  Circle  and  its  Paris.  - 

In  the  annexed  circle  ABCD,  the 
arch  line  ABCD  is  called  the  periphe- 
ryy  the  length  of  which  is  called  the 
circumference :  Any  line,  as  DB  or  AC, 
passing  through  the  centre  E,  cuts  the 
circle  into  two  equal  parts,  called  se- 
micircles, or  half  circles  ;  and  such 
lines  are  called  diameters  of  the  cir- 
cle :  If  two  diameters  be  drawn  through 
a  circle,  at  right  angles  to  each  other, 
then,  the  four  equal  divisions  of  the 
oircle  are  called  quadrants  :  half  the  diameter  as  EB,  is  called  the 
radiuSf  or  semidiameter. 

Art.  12.  The  Diameter  of  a  Circle  being  given,  to  find  the  Circum- 
ference.'^ 
Rule.  This  may  be  done  by  either  of  the  following  propor- 
tions in  whole  numbers,  as  7  is  to  22,  or  more  exactly,  as  113  is  to 
355  ;  or  in  decimals,  as  1  is  to  3- 14 159  ;  so  is  the  diameter  of  a 
circle  to  the  circumference. 

*  J^ote.     1.  If  the  diameter  of  any  circle 
be  \  "^"jtipjied  )  y^    <  3-14159,  the  product  )  -^  ^^  eircuinference. 
^divided       S"yX    -31831,  the  quotient  S 

2.  If  the  diameter  of  any  circle 

be  \  «»«}tiplied  )  .     (    -886227,  the  product  )  .     ^^  ^-^^^  ^f  ^,-,  equal  square, 
divided       5^^1-128379,  the  quotients 

3.  If  the  diameter  of  any  circle 

be  \  n^'^l^lied  \  ,     (  -866024,  the  product  )  is  the  side  of  the  equilateral 
\  divided       SI  '1547,      the  quotient  S  triangle  inscribed. 

4.  If  the  diameter  of  any  circle 

be  \  "multiplied  \ ,     S    -707016,  the  product   \  is  the^ide  of  the  square 

I  divided        \  "^  \  1-414213,  the  quoUent  \  inscribed. 

5.  If  the  square  of  the  diameter  of  any  circle 

C  multiplied  ) .     \    -785398,  the  product  ^  •   ,, 
^^  Idivided       I  ^y  ]  1-273241;  the  quotient  \  ''  ^^"^  ^''''- 

6 .  If  the  circumference  of  any  ci rcle 

7.  It  the  circumference  of  any  circle 

,  ^  $  multiplied  )  ,     J    -282094,  the  product   )  is  the  side  of  tlie 
'  Idivided       I    ^  (  3-544907,  the  quotient  \  square  equal. 

8.  If  the  circumference  of  any  circle 

^     S  multiplied  ) ,     \    -2756646,  the  product  )  is  the  side  of  the  equilateral 
''^  idivided       \  ^^  }  3-6275939,  the  quotient  S  triangle  inscribed. 

9.  If  the  circumference  of  any  circle 
j^    C  multiplied  7  ^    C   -225079,  the  product  7  is  the  side  of  the 
'^  idivided      5    ^  14-442877,  the  quotient  J  square  insci'ibed. 
10.  If  the  square  of  the  circumference  of  any  circle 
,     C multiplied  ) ,     S      -079577525,  the  product  (  ■    ,,^  ^^„^ 
'''  idivided       \  ^y  \  12-56636217,    the  quotient  \ ''  ^^'^  ^""^- 
n.  If  the  area  of  any  circle 
)  multiplied  )  ,     J  1-273241,  the  product  )  is  the  square  of 
^^  I  divided       \    ^  (    -785398,  the  quotient  \  the  diameter. 


452i  MENSURATION  OF  SUPERFICIES 

ExAMP.  A  circle  whose  diameter  is  12,  to  find  the  circumference. 
As  7  :  22  ::  12  As  113  :  356  ::  12  As  1  :  3-14159  ::  12 

12  !2  12 


7)264(37-71  =^  cir-  >  113)4260(37-699  cir.  37-69908  cir. 

21  cumference.  S         339 


64  870 

49  791 

5C  790 

49  678 

10  1120 

7  1017 

3  103 

J^ote.  3-14159  may  be  contracted  to  31416  without  any  sensible 
difference. 

Art.  13.  The  Circumference  of  a  Circle  being  given,  to  find  the  Di- 
ameter. 

RuLft.  As  22  is  to  7;  or  355  to  113;  or  as  1  to  -31831,  so  is 
the  circumference  of  a  circle  to  the  diameter. 

ExAMP.  The  circumference  of  a  circle  being  326,  to  find  the 
diameter. 

1 2.  If  the  area  of  any  circle 
,     J  multiplied?,     J  12-56636217,    the  product  ?  is  the  square  of  the 
I  divided       S    ^  f      -079577525,  the  quotient  I  circumference. 

13.  When  the  diameter  of  1  circle  is  1,  and  the  diameter  of  another  is  2,  th« 
circumference  of  the  first  is  equal  to  the  area  of  the  second,=3' 141592. 

14.  If  the  circumference  be  4,  the  diameter  and  area  are  equal,= 1-273241. 

15.  If  the  diameter  be  4,  the  circumference  and  area  are  equal,=  12-566368. 
Hence,  because  circles  are  the  most  capacious  of  all  figures,  if  ihe/burth  part 

of  a  circle  be  squared^  it  will  not  be  equal  to  the  area  of  that  circle  (as  is  common- 
ly supposed)  although  the  four  sides  added  together  are  equal  to  the  circumfa- 
rence  of  that  circle. 

In  a  circle  whose  diameter  is  24,  circumference  75*4,  and  area  452-4,  the  fourth. 
part  of  the  circumference  is  18'85,  the  square  of  which  is  only  355*3225,  that  i?, 
97-0775  less  than  the  truth  :  and  the  larger  the  circle  is,  the  greater  will  the  er- 
rour  be. 

For  further  proof  of  this  matter  ;  If  a  cylindrical  pint,  beer  measure,  whose 
content  is  35-25  cubicle  inches,  be  beaten  into  a  perfectly  square  form,  it  will  con- 
tain only  28-902  cubick  inches,  which  is  less  than  the  truth  by  6-3484+  ;  the  area 
of  the  circle  is  8-7615859288,  and  the  area  ofthe  square  only  6-8813320653076624. 

Hence  appears  the  reason,  why  taking  the  fourth  part  of  the  girth  in  measur- 
ing a  cylinder  (or  a  round  stick  of  timber)  is  false. 

16.  If  the  diameter  of  one  circle  be  doable  to  that  of  another,  the  area  ofthe 
first  circle  will  be  four  times  the  area  of  the  second,  because  the  areas  of  circles 
are  as  the  squares  of  their  diameters ;  see  Art.  15. 


^ 


AND  SOLIDS. 

453 

22  :  7  ::  326 

365:113:: 

326 

1  :  -31831  ::  326 

7 

326 
•72  diam.     678 

326 

22)2282(103 

190986 

22 

226 

63662 

82 

339 

95493 

66 

355)36838(103-76  diam. 

103-76906  =  di- 

,  

355 

ameter.     This 

160 

. 

proportion     is 

154 

1338 

the  most  accu- 

1065 

rate. 

60 

44 

2730 

— 

2485 

16  

245 

Art.  14.     To  find  the  Area  of  a  Circle. 

Rule.     Multiply  half  the  diameter  by  half  the  circumference 
and  the  product  is  the  area. 

If  the  diameter  be  given,  find  the  circumference  by  Art.  12. 
If  the  circumference  be  given,  find  the  diameter  by  Art.  13. 
ExAMP.     A  circle  whose  diameter  is  12,  and  circumference  is 
37-7,  given,  to  find  the  area? 

18-85=half  the  circumference. 
6=half  the  diameter. 


113-10=ar€a  of  the  given  circle. 
JVote.     A  circular  ring  is  the  figure  contained  between  the  peri- 
pheries of  two  concentric  circles.     Hence,  the  area  of  a  circular 
ring  must  be  the  difference  of  the  areas  of  the  two  circles. 

Art.   15.     The  Diameter  being  given  to  find  the  Area  of  a  Circle 
without  finding  the  Circumference. 

RuLF.     Multiply  the  square  of  the  diameter  by  -7854,*  and  the 
product  will  be  the  area  of  the  circle,  whose  diameter  was  given. 
ExAMP.     The  diameter  of  a  circle  being  12,  to  find  the  area  ? 
•7854 
12x12=      144 


31416 
31416 

7854 

1130976~area. 

*  When  the  diameter  is  1,  the  area  is  found  to  be  -7854,  and  as  the  areas  of 
■circles  are  as  the  squares  of  their  diameter?,  the  rule  is  evident. 


# 


454  MENSURATION  OF  SUPERFICIES 

By  the  Sliding  Rule. 

Set  1  on  A  to  the  diameter  on  B,  then  find  -7854  (which  ex- 
presses the  area  of  a  circle  whose  diameter  is  1)  on  A,  against 
which  on  B  is  a  4th  number,  then  find  this  4th  number  on  A, 
against  which  on  B  is  the  area. 

By  Gunter. 

The  extent  from  1  to  the  length  of  the  diameter  reaches  from 
•7864  to  a  4th  number,  and  from  that  4th  number  to  the  area. 

Art.  16.     The  Circumference  of  a  Circle  being  given,  to  find  the 
Area  without  finding  the  Diameter. 

Rule.  Multiply  the  square  of  the  circumference  by  -07958, 
and  the  product  will  be  the  area  of  the  circle. 

ExAMP.  The  circumference  of  a  circle  being  37*7,  to  find  the 
area.  1421-29 

37'7  -07958 

37-7 


1137032 

710G45 

1279161 

994903 

2639 
2639 
1131 

1421'29™square.        1131062582=area  of  the  circle. 

Art.   17.      The  Dimensions  of  any  of  the  parts  of  a  Circle  being 
given,  to  iind  the  side  of  a  Square  equal-  to  the  Circle. 

Rule.  If  the  area  of  the  circle  be  given,  extract  the  square  root 
of  the  area,  which  will  be  the  side  of  a  square  equal  to  the  circle  : 
If  the  diameter  or  circumference  be  given,  find  the  area  by  Art.  15 
or  16,  and  then  extract  the  square  root,  as  before.  And  this  is  a 
general  rule  to  find  the  side  of  a  square  equal  to  any  superficial  fig- 
ure, regular  or  irregular  :  for  the  square  root  of  the  area  of  any 
figure  whatever,  is  the  side  of  a  square  equal  to  the  given  figure. 
But  with  regard  to  circles,  if  the  diameter  be  given  ;  multiply  it 
by  -886,  and  the  product  will  be  the  side  of  an  equal  square  :  or,  as 
1*3'545  is  to  12,  or  1354  to  1200:  so  is  the  diameter  of  a  circle  to 
the  side  of  a  square  equal  to  the  given  circle.  And,  if  the  cir- 
cumference be  given,  multiply  it  by  -282  for  the  side  of  an  equal 
square.  Or,  divide  it  by  3-515,  and  the  quotient  »vill  be  the  side 
of  an  equal  square. 

ExAiiP.   1.  Exam  p.  2. 

Let  the  diameter  of  a  circle  be  The  circumference  being  37  7 
12,  to  find  the  side  of  a  square  to  find  the  side  of  an  equal 
equal  to  the  circle  ?  square  ? 

•886xl2=10-632=side  of  the       37-7X'282=10-63I  =side      of 
spuare.  the  square. 

Or, as  13-545:  12::  12:  10  631       Or,  37'7-~3-545=10-63i, 
—the  side. 


AND  SOLIDS. 


453 


Art.  18.     The  Area  of  a  Circle  being  given,  iojind  the  Diameter. 

KuLE.  Multiply  the  given  area  by  1-2732,  and  the  product  will 
be  the  square  of  the  diameter  ;  then,  extracting  the  square  root  of 
the  pr(Kiuct,  you  will  have  the  diameter.* 

ExAMP.  The  area  of  a  circle  being  113  09,  to  find  the  diameter. 


1-2732 
11309 

143-986188(11*999=^ 
1 

114588 
381960 
12732 
12732 

21)43 
2^1 

229)2298 
2061 

143-986188 

2389)23761 
21501 

23989)226088 
215901 

10187  remainder. 

Art.  19.     Tlie  Area  of  a  Circle  being  given,  to  find  the  circumference. 
Rule.    Multiply  the  given  area  by  12-566,  and  extract  the  square 
root  of  the  product,  which  root  will  be  the  circumference  required. 
ExAMP.     The  area  of  a  circle  being  113  03  to  find  the  circumfe- 
rence. 


12-566 
11303 

1420'3349(37-68=circumferencc, 
9 

37698 
376980 
12566 
12566 

67)520 
469 

746)5133 
4476 

1420-33498 

7528)65749 
60224 

5525  remainder. 

Art.  20.    The  Side  of  a  Square  being  given,  to  find  the  Diameter  of  a 
circle  equal  to  the  Square,  whose  Side  is  given. 

Rule.  Multiply  the  given  side  by  1128,  and  the  product  will 
be  the  diameter  of  a  circle,  whose  area  is  equal  to  the  area  of  the 

*  As  the  area  of  a  circle,  whose  diameter  is  1,  is  -7854,  the  area  divided  by 
•7{{54  must  give  the  square  of  the  diameter ;  but  a?  l-273'i2  is  the  reciprocal  ©f 
'7854,  the  rule  is  evident. 


436  MENSURATION  OF  SUPERFICIES 

given  square.  Or,  if  the  side  of  the  square  be  divided  by  -886,  the 
quotient  will  be  the  diameter.  Or,  as  12  to  13-34,  so  is  the  side  of 
any  square  to  the  diameter  of  an  equal  circle. 

ExAMP.  The  side  of  a  square  being  10*635,  to  find  the  diameter 
«f  a  circle  equal  to  that  square  ? 

10-636XM28=12  nearly.     Or,  10-635-~-886=12=diameter. 
Or,  as  12  :  13-54  ::  10  635  :  12  nearly* 

Art.  21.   The  Side  of  a  Square  being  given,  to  find  the  circumference 
of  a  Circle  equal  to  the  given  Squaie. 

Rule.  Multiply  the  given  side  by  3-645  and  the  product  will 
be  the  circumference  required.  Or,  divide  it  by  282,  and  the  quo- 
tient will  be  the  circumference. 

ExAMp.  The  side  of  a  square  being  10-631,  to  find  the  circum- 
ference of  a  circle  equal  to  that  square. 

10-631x3  645=37'686=circum.  Or, -282)  10-631  (37-698  circum. 

Art.  22,     To  find  the  Area  of  a  Semicircle,  the  Diameter  being  given. 

Rule.  Find  the  area  of  the  circle  by  Art.  15,  and  take  the  half 
of  it. 

In  the  same  manner  may  the  area  of  a  quadrant,  or  a  quarter  of 
a  circle,  be  found,  by  taking  a  fourth  part  of  the  area  of  the  whole 
circle. 

But  with  regard  to  measuring  a  sector,  or  a  segment  of  a  circle, 
it  will  be  necessary  first  to  show  how  to  find  the  length  of  the  arch 
line  of  a  sector,  and  the  diameter  of  the  circle  to  a  given  segment. 

Art.  23.     A  Segment  of  a  Circle  being  given,  to  find  the  length  of  the 

Arch  Line. 

Rule.  Divide  the  segment  into  two  equal  parts  ;  then  measure 
the  chord  of  the  half  arch,  from  the  double  of  which  subtract  the 
chord  of  the  whole  segment ;  and  one  third  of  that  difference,  be- 
ing sdded  to  the  double  of  the  chord  of  the  half  arch,  will  give  the 
length  of  the  arch  line. 

ExAMP.  In  the  segment  ABCD,                            jj 
the  whole  chord  ADC  is  216,  and                ^x-'^t!^^^^^'^'^^^^'*^*^ 
the  chord  AB  or  BC  126,  to  find           y^^;f       j  ^N^\ 
the  arch  line  ABC.                              /^^^          '          ^^*^^\ 
126=chord  AB  or  BC.  A^:L _J JS^ 

252=double.  252=double  of  AB. 

216=ADC,  to  be  subtracted.  12=^  difference  added. 

3)36=difrerence.  264=Ienglh  of  the  arch  ABC. 

12x=^  difference . 


AND  SOLIDS. 


457 


Art.  24.     The  Chord  and  versed  Sine  of  a  Segment  being  given,  to  find 
the  Diameter  of  a  Circle. 
Rule.     Multiply  half  the  chord  by  itself,  and  divide  the  product 
by  the  versed  sine  ;  then  add  the  quotient  to  the  veHed  sine,  and 
the  sum  will  be  the  diameter  of  the  circle. 


B 


Example.  In  the  segment 
ABCD,  the  chord  AC  is  1869-5, 
and  the  versed  sine  13D  4235, 
to  find  the  diameter. 

{  half  the 


934-75 
934-75 


I  chord  AC 


467375 
664325 
373900 
280425 
841275 

423 

"•5)873757-5625(2063  1  = 
8470       423-6= 

=DE 
=BD, 

"E 


26757 
25410 


2486-6=diameter  BDF. 


13475 
12705 


1 


7706 
4235 


3471 

Art.  25.     To  measure  n  Sector. 
Definition.     A  sector  is  a  part  of  a  circle,  contained  between  an 
arch  line,  and  two  radii  or  semidiameters  of  the  circle.  ^""''^S  >'^'< 
Rule.     Find  the  length  of  half  the  arch  by  Art.  23  :  Then  mul- 
tiply this  by  the  radius  or  semidiameter,    and  the  product  will  be 
the  area. 

ExAMP.  1.  In  the  sector  ABCD, 
given  the  radius  AD  or  DC  72 
feet,  the  chord  AC=126  feet,  and 
the  chord  AB  or  BC=70,  to  find 
the  area  of  the  sector. 
First. 
70=chord  AB  or  BG. 
2 
—~  Carried  over. 

K  3 


458 


MENSURATION  OF  SUPERFICIES 


140         Brought  over. 
126^AC,  subtract. 

3)14 

4'GC 
140 


Secondly. 
72-33=halfthe  arch. 
72=railius. 

14466 
60631 


144'16=length   of    the    arch    5207-76=area. 

[ABC,by  Art.  23. 

72-33 

E.YAMP.  2.  In  the  sector  ABCD, 
greater  than  a  semicircle,  given  the 
radius  AE  or  ED=112,  the  chord 
BD  (of  half  the  arch  ABD)==204, 
and  the  chord  BC  (of  half  the  arch 
BCD)=120,  to  find  the  area  of  the 
sector. 

120=::BC. 

!  2 


240 

204  subtract. 

3)36 

12 
240  Add. 


252=half  the  arch  ABD. 
112=radius. 


S04 
252 

262 


Che  arch 
Art.  23. 


28224=area  of  the  sector. 


ot^s—  ^i-engthof 
-^^~  I  BCD,  by  . 

Art.  26.     To  find  the  Area  of  a  Segment  of  a  Circle. 

Definition.  A  segment  of  a  circle  is  any  part  of  a  circle  cut  off 
by  a  right  line  drawn  across  the  circle,  which  does  not  pass  through 
the  centre,  and  is  always  greater  or  less  than  a  semicircle. 

ExAMP.  1.     To  find  the  area  of  the  segment  ABC,  whose  chorrt 
AC  is  172,  tlie  chord  of  half  the  arch  ABC,  viz.  BC=104,  and  tlit> 
versed  sine  BD=68-48. 

RuL£.  By  Art.  23,  find  the  length  of 
the  arch  line  ABC,  and  by  Art.  24,  the 
diameter  FB  ;  (hen  multiply  half  the  a 
chord  of  the  arch  ABC  by  half  the  diam- 
eter, and  the  product  will  be  the  area  of 
the  sector  ABCE  :  then  find  the  area  of 
the  triangle  AEC,  whose  base  AC  is  172, 
and  perpendicular  heiglU  34,  found  by 
subtracting  the  versed  sine  BD  from  half 
the  diameter;  and  the  area  of  Ibe  trian- 
gle AEC,  being  subtracted  from  the  area 
of  the  sector  ABCE,  will  leave  the  area  of  the  segment  ABC 


AND  SOLIDS. 


4i9 


104=BC. 

2 

208 

172= AC,  subtract. 

3)36 


86=halfADC. 

86 


516 

688 


58-48)7396-00(126-47=DEF 
5848  ' 


58'48=BD,  add. 


12 
208  add> 

220=arch  line  ABC. 

110=^halfarch. 

92*475=radiu8. 
110 


1 5480       184-95=diameter  BF. 
11696       

37840  I      diameter. 

35088 


924750 

92475 


27520 
23392 

41280 
40936 


10172-25=areaorthe  sector.  344 

86=half  the  base=AD.        101 72  25=area  of  the  sector. 
34=perpendicular  DE.  2924      =area  of  the  triangle. 

7248  25=area  of  the  segment. 


2924=area  of  the  triangle. 

ExAMP.  2.     In  the  segment  ABCD  greater  than  a  semicircle, giv- 


en the  chord  of  the  whole  segment  AD= 136,  the  chord  AC  of  half 


the  arch  ACD=146,  the  chord  AB 
or  BC  one  fourth  of  the  arch  ACD 
=86,  and  the  radius  AE  or  ED= 
80,  to  find  the  area  of  the  segment 
ABCD. 

First  find  the  area  of  the  sector 
ABCDE,  by  Art.  25,  at  the  second 
Example  ;  then  find  the  area  of  the 
triangle  AED,  by  Art.  6,  and,  ad- 
ding the  area  of  the  triangle  to  the 
area  of  the  sector,  you  will  have 
the  area  of  the  segment. 
86=chord  AB. 


172 

146=chord  AC,  subtract. 


736 


8-66G 
172        =double  of  AB,  add. 


180-666=arch  line  ABC. 
80=radius. 


3)20 


IM53»280=area  of  the  sector 
Carried  over* 


460 


MENSURATION  OP  .SUPERFFCIES 


Brought  over. 
68=half  the  base  AD. 
42=perpendicular  E  136. 


2856=area  of  the  triangle  AED. 
14453'28=area  of  the    sector, 

[add. 

17309'28=area  of  the  segment. 


136 
272 

Note  1.  The  area  of  a  Lune  or  Crescent,  is  calculated  by  the 
preceding  rule.  A  Lune  is  a  figure  made  by  two  circular  arcg, 
which  intersect  each  other,  as  ACBD. 
The  area  of  the  Lune  is  the  difference 
of  the  two  segments,  which  are  contain- 
ed by  the  arcs  and  the  chord.  Thus  the 
difference  of  the  segments  ACBE  and 
ADBE  is  the  area  of  the  crescent  ACBD. 

Note  2>     A  Circular  Zone  is  a  figure  con- 
tained between  two  parallel  chords.     If  the 
chords  be  equal,  it  is  called  a  middle  zone,  as  A 
ABCD.     The  area  of  a  zone  is  evidently  the 
difference  between  the  area  of  the  circle  and  D 
the  areas  of  the  two  segments. 

Art.  27.     To  find  the  Area  of  an  Ellipsis, 

Definition.  An  ellipsis,  or  oval,  is  a  curve  which  returns  into  it- 
self like  a  circle,  but  has  two  diameters,  one  longer  than  the  oth- 
er, the  longest  of  which  is  called  the  transverse,  and  the  shortest 
the  conjugate  diameter. 

Rule.  Multiply  the  two  diameters  of  the  ellipsis  together  ;  then 
multiplying  the  product  by  -7854,  this  last  product  will  be  the  area 
of  the  ellipsis. 

ExAMP.  In  the  ellipsis  ABCD,  the 
transverse  diameter  AC  is  88,  and  the 
conjugate  diameter  BD  is  72,  to  find 
the  area, 

88 
72 


176 
C16 

C336 
7854 


25344 
31680 
50688 
44352 


The  conteftt  is  found  by  the  sliding  rule 
and  Gunter,  in  the  same  way  as  the  circle, 
only  using  the  product  of  the  two  diam- 
eters as  the  square  of  the  diameter  of  a 
circle. 


4976-2944=area. 

Mensuration  of  Superficies  is  easily  applied  to  Surveying :  lhu«, 
take  the  angles  of  the  plot  with  a  good  compass,  then  Dpkeasure  the 


AND  SOUDS.  461 

sides  with  Gunter's  chain,  which  note  down  in  links  (or  chains  and 
links,  which  is  done  by  separating  the  two  right  hand  figures  of  your 
links  by  a  comma,  your  chain  being  100  links)  then  cast  up  the  con- 
tents, according  to  the  rule  of  the  figure,  cutting  off  the  five  right 
hand  figures  of  the  product,  and  those  at  the  left  hand,  if  any,  are 
acres ;  then  multiply  the  five  figures  cut  off,  by  4,  by  40,  and  by 
272i,  cutting  off  as  before,  and  those  at  the  left  hand,  will  be  roods, 
poles,  and  feet,  respectively. 

Section  II.     Of  Solids. 

Solids  are  measured  by  the  solid  inch,  foot,  or  yard,  kc.  1728  of 
these  inches,  that  is  12x12x12,  make  one  cubick  or  solid  foot. 

The  solid  content  of  every  body  is  found  by  rules  adapted  to 
their  particular  figures. 

Art.  28.     To  measure  a  Cube.* 

Definition.  A  cube  is  a  solid  of  six  equal  sides,  each  of  which  is 
an  exact  square. 

*  Here  follows  a  Table  of  the  Proportions,  which  the  following  Solids  hav;-^ 
to  the  Cube  and  Cylinder,  having  the  same  Base  and  Altitude.     Solid  Inches. 

1.  A  Cube  whose  side  is  12  inches,  contains  1728 

2.  A  Prisw,  having  an  equilateral  triangle,  whose  side  is  12  )     j.'q^.oi 
im;hes  from  its  Base^  and  its  Altitude  12  inches,  contains  \     <o4'..4 

3.  A  Square  Pyramid,  whose  height  and  the  side  of  its  base,  are  ?     ^ 
eacli  12  inches,  is  h  of  the  above  cube,  and  therefore  contains         \ 

4.  A  Triangular  Pyramid,  whose  height  and  side  of  its  triangu-  \ 
]jjr  base  are  each  12  inches,  is  near  j.  of  the  cube,  and  contains      \ 

5y  A  Cylinder,  whose  diameter  and  Jieight  are  each  12  inches,  > 
is  'i-1  of  the  above  cube,  and  contains  \ 

6.  A  Sphere  or  Globe,  whose  axis  or  diameter  is  12  inches,  equal ) 
to  the  side  of  the  cube,  is  11  of  it,  and  contains  > 

7.  A  Cone,  whose  base  and  altitude  are  each  12  inches,  equal  ) 
to  the  side  of  the  cube,  is  JL.  of  it,  and  contains  ^ 

8.  A  ParaJboHck  Conoid,  -whose  diameter  at  the  base  and  height,  )     p'~?.ro«5 
are  each  12  inches,  being  ^  its  circumscribing  cylinder,  contains      s,        '  ^'^^"^ 

9.  A  Hyperbolick  Conoid,  whose  height,  and  diameter  at  the  ) 

base,  are  each  12  inches,  is   5    of  its  circumscribing  cylinder,  and  V     565'49 
contains  ) 

10.  A  Parabolick  Spindlc,-whose  height  and  middle  diameter  are  }     ^-o'^.oo^ 
each  12  inches,  is  JL  of  its  circumscribing  cylinder,  and  contains    <>     '""^  ^* 

Hence  arises  a  different  method  of  finding  their  contents. 

General  Rule.  If  the  base  of  the  solid,  whose  contents  you  would  find,  I  e 
rectilinear,  consider  it  as  Parallehpipedon ;  if  curved,  as  a  Cylinder,  and  find 
the  content  accordingly  :  then  take  such  a  part  of  the  content,  thus  found,  as  is 
specified  in  the  preceding  Table,  which  if  the  parts  be  taken  in  incher-,  will  bn 
the  solid  content  of  the  given  Bgure,  in  inches,  wliich,  divided  by  1728,  will  givr 
the  cubick  feet. 

ExAMP.  1.  There  is  a  triangular  prism,  the  side  of  whose  base  is  48  inche-, 
and  whose  perpendicular  height  is  108  inches  :  what  is  its  solid  content? 

The  base  being  right  lined,  I  consider  it  as  a  paraUelopipedon,  the  side  cf 
whose  base  it  4o  inches,  and  whose  length  is  108  inches,  and  as  784*24  is  con  - 


249-41:^. 

1357-17 

904-78 

452-38829 

462  MENSURATION  OF  SUPERFICIES 

The  solid  foot  is  composed  of  1728  inches ;  for  a  8011(1,  that  is 
1  foot,  or  12  inches  every  way,  that  is  12x12x12,  contains  172S 
inches. 

Rule.  Multiply  the  side  by  itself  and  that  product  by  the  same 
side,  and  this  last  product  will  be  the  solid  content  of  the  cube.t 

ExABip.  The  side  of  a  cube  AB,  being  18 
inches,  or  1  foot  and  6  inches,  to  find  the 
content  ? 

1  foot  6  inches=l'5  foot.  18  inches. 

15  18 

75  144 

15  18 

Carried  over.        2-25  324 

tained  2'20340712  times  in  a  cubick  foot ;  2-20340712  is  a  divisor,  to  divide  the 
content  of  the  parallelopipedon  by;  therefore  48X48X108^2*203407 12= 
112930-56  solid  inches=65-353  solid  feet. 

Had  the  dimensions  been  given  in  feet,  it  would  have  been  4x4X9-t- 
2-203407 12=65-353  feet. 

ExAMT.  2.  There  is  a  square  pyramid,  whose  height  is  12  feet,  and  the  side 
of  whose  base  is  3-5  feet ;  what  is  its  content  ? 

3-5  X  3-5  X  12-^3=49  feet,  Ans. 
ExAMP.  3.  There  is  a  triangular  pyramid,  whose  height  is  15  feet,  and  the 
side  of  whose  base  is  5  feet :  what  is  its  content  ? 

5 X5 X  15-^-7=53-57  feet,  Ans. 
ExAMP.  4.  There  is  a  cylinder  whose  diameter  is  2-5  feet,  and  whose  length 
is  24  feet ;  what  is  its  content  ? 

Here,  the  diameter  is  to  be  considered  as  the  side  of  the  base  of  a  parallelo- 
pipedon.    Therefore,  2-5  X2-5X24Xll-M4=l  17-857  feet,  Ans. 

Ex  AMP.  5.  There  is  a  spherical  balloon,  whose  diameter  is  50  feet ;  how  many 
eubick  feet  of  air  does  it  contain  ? 

Here,  the  diameter  is  to  be  considered  as  the  side  of  a  cube.     Therefore, 

50x60X50x11-^21=65476-19  feet,  Ans. 
ExAMP.  6.  There  is  a  cone,  whose  height  is  15  feet,  and  the  diameter  of 
whose  base  is  5  feet ;  what  is  its  content? 

Here,  the  diameter  of  tlie  base  is  to  be  considered  as  the  side  of  the  base  of  a 
parallelopipedon,  and  its  height,  as  the  length.     Therefore, 

5X5X15X5-^19=98^84  feet,  Ans. 
ExAMP.  7.  There  is  a  parabolick  conoid,  whose  diameter  at  the  base  is  2-9 
feet,  and  whose  height  is  6  feet ;  what  is  the  content  ? 

This  solid  being  ^  of  a  cylinder  ;  we  must  first  find  the  content  as  of  that  of 
a  cylinder,  and  then  halve  it.     Therefore, 

2-9  X2-9  X6  X  11-^14=39-647,  and  39-647-^2=19-823,  Ans. 
ExAMP.  8.  There  is  a  hyperbolick  conoid,  whose  diameter  at  the  base  is  2-9 
feet,  and  whose  height  is  6  feet ;  what  is  the  content  ? 

Fii-st,  find  the  <?ontent  of  a  cylinder. 
2-9 X2-9X6X11-M 4=39-647,  and  39-647 X/_=16-519  feet,  Ans. 
ExAMP.  9.  There  is  a  parabolick  spindle,  whose  middle  dian^eter  is  2-9  feet, 
and  whose  length  is  6  feet ;  required  the  content  ? 

First,  find  the  content  of  a  cylinder. 
2-9x2-9X6 Xll-M4=39-647,  and  39^647 Xj-8^=21' 145  feet,  Ans. 

f  Multiplying  a  side  by  itself,  or  squaring  a  side,  gives  the  area  of  the  base, 
or  the  number  of  square  inches,  feet,  &c.  in  the  base ;  whence  one  inch,  foot,  &c. 
in  height  would  give  as  many  solid  inches,  feet,  &c.  as  there  are  squares  in  the 
base  ;  two  inches,  &c.  in  height,  twice  as  many,  and  so  ori,  and  is  the  rule,  when 
the  sides  are  equal  to  each  other.  In  tlie  same  way,  the  rule  for  the  content  cf 
♦he  Parallelopipedon  is  proved. 

/ 


AND  SOLIDS 


^' 


•'   Brought  up.     2  25 
1-5 


1125 

225 


324 
18 

2592  ♦ 
324 


3  375      1728)5832(3-375 
5184 

In  this  operation,  the 

inches  are  changed  into  6480 

the   decimal  parts  of  a  5184 

foot.  

12960. 
12096 


8640 
8640 


I  have  done  this  two  different  ways,  that  the  learner  may  sec 
they  come  out  the  same.  The  content  in  inches  is  5832,  which 
being  divided  by  1728,  the  inches  in  a  solid  foot,  and  the  division 
continued  by  annexing  cyphers,  it  comes  out  the  same  as  the  deci- 
jenal  operation. 

Note.  The  area  of  the  surface,  or  superficial  content  of  the 
cube  and  parallelopipedon  is  found  by  adding  the  areas  of  the  sev- 
eral quadrilateral  figures  which  compose  them. 

Art.  29.  To  measure  a  Parallelopipedon. 
Definition.     A  parallelopipedon  is  a  solid  of  three  dimensions, 
length,  breadth  and  thickness  ;  as  a  piece  of  timber  exactly  squar- 
ed, whose  length  is  more  than  the  breadth  and  thickness.     The 
ends  are  called  bases,  which  are  equal. 

Rule.  Find  the  area  of  the  base,  then  multiply  that  by  the 
length,  and  it  will  give  the  solid  content. 

ExAMP.  1.  The  side  AB  is  1-75  foot,  and  the  length  AD  9*5  feet, 
to  find  the  solid  content  ? 
1-75—1  foot,  .9  inches. 
1-75 


875 
•    1225 
175 

3  0625=area  of  base 
9-5 

153125 

275625 


9/5 


ExAMP.  2.  A  vessel  3-5  feet  each 
side  within,  and  5  feet  deep,  to  find 
the  content  ? 


2909-'^75^'Sclid  content. 


3-5 

.3-5 

175 
105 


12-25 


61'25si^the  content. 


464 


MENSURATION  OF  SUPERFICIES 


If  a  piece  of  timber,  or  any  other  thing,  be  of  an  equal  bigness 
through  its  whole  length,  though  there  be  a  difference  between  the 
breadth  and  thickness,  if  the  breadth  and  thickness  are  multiplied 
together,  and  that  product  multiplied  by  the  length,  this  last  pro- 
duct will  be  the  solid  content. 

ExAMP.  3.  A  piece  of  timber  being  1  foot  and  6  inches,  or  18 
inches  broad,  9  inches  thick,  and  9  feet  6  inches,  or  114  inches 
long,  to  find  the  content  ? 

1  foot   6  inches=  1-5  foot  Breadth  =18  inches. 

9  inches='75  foot.  Depth    =  9  inches. 


75 
105 


162 
Length=114  inches. 


M25 
9feet6inches=    9-5 


5625 
10125 

10  6875=content. 

In  this  operation  thfe  inches 
are  changed  into  the  decimal 
fractions  of  a  foot. 


648 

162 

162 

1728)18468(10-6875=conteni, 
1728  as  before- 


11880 
10368 


15120 
13824 

12960 
12096 


8640 
8640 

Notct,  VVhen  the  end  is  given  in  inches  and  the  length  in  feet, 
find  the  area  at  the  end  in  inches,  multiply  that  by  the  length  m 
feet,  and  divide  this  product  by  144  (the  square  inches  in  a  foot) 
and  the  quotient  will  be  the  feet. 


Take  the  last  example. 
Foot. 

1*5  =18  inches. 
'75=  9  inches. 

162  area  in  inches. 
9-5  feet=length. 


By  the  sliding  Rule. 
Set  12  inches  on  the  girt  line  D 
to  the  side  of  the  square  end  on  C, 
then,  against  the  length  on  D,  you 
will  have  the  answer  on  C. 


810 
1458 


144)1 539(10'68' 


By  Gunier. 

Extend  the    compasses  from    li* 

inches  to  the  length  of  the  side  of 

the    square    end  ;     that     distance, 

twice  turned  over  from  the  length, 

content,  will  reach  to  the  contetit. 


AND  SOLIDS. 


465 


When  the  side  of  a  square  solid  is  given,  in  inches,  to  find  how 
niuch  in  length  will  make  a  foot  solid. 

Rule.  As  the  given  side  is  to  12,  so  is  12  to  a  fourth  number, 
and  so  is  that  fourth  number  to  its  required  length.  Or  divide  1728 
by  the  area  at  the  end,  and  the  quotient  will  be  the  length  making 
a  solid  foot. 

If  the  given  side  is  in  foot  measure,  then, 

Rule.  As  the  given  side  is  to  1  ;  so  is  1  to  a  fourth  number, 
and  so  is  that  fourth  number  to  the  required  length. 

When  two  sides  of  an  equal  square  solid  (that  is,  of  unequal 
breadth)  are  given,  to  find  what  length  will  make  any  number  of 
solid  feet. 

Rule.  Multiply  the  proposed  number  of  feet  by  144 :  divide 
that  product  by  the  product  of  the  breadth  and  depth,  and  the 
quotient  will  be  the  length  required. 

Art.  30.    To  measure  a  Cylinder. 

t)efinition.  A  cylinder  is  a  round  body,  whose  bases  are  circles, 
like  a  round  column,  or  a  rolling  stone  of  a  garden. 

Rule.  The  diameter  of  the  base  being  given,  find  the  area  of 
the  end  by  Art.  15,  then,  multiplying  the  area  of  the  base  by  the 
length,  that  product  will  be  the  content  of  the  cylinder. 


ExAMP.  The  diameter  of 
the  base  AC  being  1  foot 
and  9  inches,  and  the  length 
BD  12  feet  and  6  inches,  to 


and  toe  conte 

nt. 

I2| 

1-75= 
1-75 

875 
1225 
175 

=diam. 

of  the  base. 

2-405=area  of  the  base 
12-5=length. 

12025 
4810 
2405 

3-0625 

•7854 

30  0625=content. 

122500 
163125 
245000 
i>  14375 

240528760=area  of  the  base. 

If  the  square  of  the  diameter  of  a  cylinder  be  multiplied  by 
•7854,  and  the  solidity  divided  by  that  product,  the  quotient  will 
be  the  length,  and.  if  the  content  be  divided  by  the  length,  the 
quotient  will  be  the  area  of  the  end,  from  which  the  iiameter  \$ 
found  by  Art,  IR. 

L  3 


iuo  x^JENSURATlON  OF  SUPERFICIES 

'j'he  learner  may,  for  his  practice,  reduce  all  the  dimensions  to 
inches,  and  find  the  solid  content  in  inches,  which  being  divided  by 
1728,  the  quotient  will  be  the  solid  content  in  feet :  or,  if  he  finds 
the  area  at  the  end  in  inches,  and  multiplies  that  by  the  length  in 
feet,  and  divides  by  144  ;  the  quotient  will  be  feet. - 

This  is  a  general  rule  for  finding  the  content  of  any  straight 
solid  body,  of  equal  bigness  from  end  to  end,  of  whatever  form  the 
bases  are  :  for,  if  the  area  of  the  base  be  multiplied  by  the  length, 
the  product  will  be  the  solid  content. 

By  the  SUditig  Rule. 

Set  13  5,  the  square  root  of  183-34  (which  is  a  guage  point 
arising  from  the  division  of  144  by  -7854)  found  on  D,  to  the  diam- 
eter found  on  C,  and  opposite  to  the  length,  on  D,  you  will  find 
the  content  on  C. 

Or,  as  42  64  is  to  the  circumference  ;  so  is  the  length  in  feet  to 
a  fourth  number,  and  so  is  that  fourth  number  to  the  answer. 

Note.  The  superficial  content  of  a  cylinder  is  found  by  multiply- 
ing the  circumference  of  one  of  the  bases  into  the  length,  and  to 
the  product  adding  the  areas  of  the  two  bases,  or  ends. 

When  the  diameter  is  given  in  inches,  to  find  what  length  will 
make  a  solid  foot. 

Rule.  As  the  given  diameter  is  to  13  631  :  so  is  12  to  a  fourth 
number,  and  so  is  that  fourth  number  to  the  required  length.  If 
the  diameter  be  given  in  foot  measure  :  Rule,  as  the  given  diame- 
ter is  to  1*128  :  so  is  1  to  a  fourth  number,  and  so  is  that  fourth 
number  to  the  required  length.  Or,  divide  1728  by  the  area  at  the 
end  in  inches,  and  the  quotient  will  be  the  required  length. 

To  find  how  much  a  Cylindrick  or  round  Tree,  that  is  equally  thick 
from  end  to  end,  will  hew  to,  when  made  square. 

Rule.  Multiply  twice  the  square  of  its  semidiameter  by  the 
length,  then  divide  the  product  by  144,  and  the  quotient  will  be 
the  answer. 

If  the  diameter  of  a  round  slick  of  timber  be  24  inches  from 
end  to  end,  and  its  length  20  feet :  how  many  solid  feet  will  it 
contain,  ivhen  hewn  square  ;  and  what  will  be  the  content  of  the 
slabs  which  reduce  it  to  a  square  ? 
12X12X2X20 
—- =40  feet,  the  solidity  when  hewn  square. 

24X24X-7854X20 

jj^ =62  8  feet,  or  2x2X*7854x20=62-8  the  total 

solidity,  whence  62-8—40=22  8  feet,  the  solidity  of  the  slabs. 

Note.  The  rule  of  workmen  for  measuring  round  timber  is  to 
multiply  the  square  of  the  quarter  girt  or  one  fourth  of  the  cir- 
cumlerence,  by  the  length.  This  rule  allows  about  one  fifth,  for 
the  bark,  waste  in  hewing,  &c.  The  example  above,  in  which  the 
diameter  of  the  cylinder  is  1  foot  9  inches,  and  the  length  12  fee* 


AND  SOLIDS, 


4&r 


6  inche?,  will  give  the  quarter  girt  1  3744  feet,  and  Hie  solid  con- 
tent is  l-3744^xl2-5=23*61  feet,  which  is  nearly  four  Mha  of 
30C625,  the  content  hy  the  accurate  rule. 

A  rule,  nearly  correct,  is  to  multiply  twice  the  square  of  one 
fifth  of  the  circumference  by  the  length.  Thus,  in  the  example,  } 
of  the  circumference  is  10995,  and  2  x  .1 -0995^x12-5=  30-22 
feet. 

Art.  31.    To  measure  a  Prism. 

Definition.  A  prism  is  a  body  with  two  equal  or  parallel  ends, 
either  square,  triangular,  or  polygonal,  and  three  or  more  sides, 
which  meet  in  parallel  lines,  running  from  the  several  angles  at 
one  end,  to  those  of  the  other. 

Rule.  Prisms  of  all  kinds,  whether  square,  triangular  or  poly- 
gonal, are  measured  by  one  general  rule,  viz.  Find  the  superficial 
content,  or  area  at  the  base  (or  end)  by  the  proper  rule  of  Sect.  K 
and  this  multiplied  by  the  length,  or  height  of  the  prism,  will  give 
the  solid  content. 

ExAMp.  The  side  of  a  stick  of  timber,  AB,  hewn 
three  square,  is  10  inches,  and  the  length,  AC,  is  12 
feet,  to  find  the  content  ? 

Side=     10  inches. 

I  Perpendicular=4-33  inches. 

43-3=area  at  the  end. 
12  feet=length. 

144)519G(3-6  feet,  content. 
432 


87{> 
864 


12 
Note.     The  superficial  content  is  found  by  adding  the  areas  of 
the  several  quadrilateral  and  triangular  figures  which  compose  it. 

Art.  32.     To  measure  a  Pyramid. 

Definition.  Solids,  which  decrease  gradually  from  the  base  till 
they  come  to  a  point,  arc  generally  called  pyramids,  and  are  of  dif- 
ferent kmds,  according  to  the  figure  of  their  bases  ;  thus,  if  it  has 
a  square  base,  it  is  called  a  square  pyramid  ;  if  a  triangular  base, 
a  triangular  pyramid  :  If  the  base  be  a  circle,  a  circular  pyramid, 
or  simply  a  cone.  The  point,  in  which  the  top  of  a  pyramid  ends, 
is  called  a  Vertex,  and  a  line  drawn  trom  the  vertex,  perpendicular 
to  the  base,  is  called  the  height  of  the  pyramid. 

Rule.  Find  the  area  of  the  base,  whether  triangular,  square, 
polygonal  or  circular,  by  the  rules  in  superficial  measure  :  then, 
multiply  this  area  by  one  third  of  tiie  height,  and  the  product  will 
be  the  solid  content  of  the  ryr^vmil 


468 


MENSURATION  OF  SUPERFICIES 


ExAMP.  !•     Id  a  triangular  pyramid,  the  height 

BE,  being  48,  and  each  side  of  the  base  13  :  the 

base  being  a  triangle,  let  the  perpendicular  height 

DE  be  1 1  ;  to  find  the  content. 

5-5=half  ED. 

13=base  AC, 

165 
55 

7]'5=area  ef  the  base. 
16=1  of  the  height  EB. 

4290 
715 


n44'0=content 

ExAMP.  2.  In  a  quadrangular  pyramid,  the  height 
JBE  being  48,  and  each  side  of  the  base  13,  to  find 
the  content. 

13 

13 


39 
13 


169=area  of  the  base. 
16=J-  of  the  height  £B. 

1014 

169 

2704==coDtent. 

ExAMP.  3.  To  measure  a  Cone^ — The  diameter 
AC  being  13,  and  the  height  BD  48,  to  find  the 
content, 

13 

13 

39 
13 

169 
•7854 

676 
845  i 

1352 
1183 


132'7326=area  of  the  base. 


AND  SOLIDS.  4«9 

Brought  up.     132-7326 

16=i  of  the  height. 

7963956 
1327326 


2123-7216=content. 
Note.  The  superficial  content  of  all  pyramids  is  fouad  by  tak- 
ing the  sum  of  the  several  areas,  which  compose  them.  That  of 
a  cone,  by  multiplying  the  circumference  of  the  base  into  half  the 
line  joining  the  vertex  and  any  point  in  that  circumference,  and 
adding  the  area  of  the  base  to  the  product. 

Art.  33.     To  measure  the  Frustum  of  a  Pyramid. 

Definition,  The  frustum  of  a  pyramid  is  what  remains  after  the 
top  is  cut  off  by  a  plane  parallel  to  the  base,  and  is  in  the  form  of 
a  log  greater  at  one  end  than  the  other,  whether  round,  or  hewn 
three  or  four  square,  &c. 

Rule,  If  it  be  the  frustum  of  a  square  pyramid,  multiply  the 
side  of  the  greater  base  by  the  side  of  the  less ;  to  this  product 
add  one  third  of  the  square  of  the  difference  of  the  sides,  and  the 
sum  will  be  the  mean  area  between  the  bases  ;  but  if  the  base  be 
any  other  regular  figure,  multiply  this  sum  by  the  proper  multipli- 
er of  its  figure  in  the  Table,  Art.  11.  and  the  product  will  be  the 
mean  area  between  the  bases  :  lastly,  multiply  this  by  the  height, 
and  it  will  give  the  height  of  the  frustum. 

ExAMP.  1.     In  the  frustum  of  a  square  pyramid  the  q 

side  of  the  greater  base  AD=16,  the  side  of  the  less,    ^  ^ 
BC=6,  and  the  height  KF=40,  to  find  the  content.        ^  ^ 
15=AD.  15 

6=BC.  6 

Prod.=;:90  9=difference. 

Add      27  9 

117  3)81?=square  of  the  difference.  A.< 

X     40  — 

27=1  of  the  square. 

4680=content. 
Or,  if  it  be  a  tapering  square  stick  of  timber,  take  the  girth  of 
it  in  the  middle  ;  square  ^  of  the  girth  (or  multiply  it  by  itself  ia 
inches)  then  say,  as  144  (inches)  to  that  product ;  so  is  the  length, 
taken  in  feet,  to  the  content  in  feet. 

ExAMP.  2.  What  is  the  content  of  a  tapering  square  stick  of 
timber,  whose  side  of  the  largest  end  is  12  inches,  of  the  least  end, 
8,  and  whose  length  is  thirty  feet. 

One  fourth  of  the  girdj  in  the  middle=10,  and  10x10=100,  the 
area  in  the  middle  ;  then,  as  144  ;  100  :;  30  feet  :  20  03  feet  the 
content. 


4T0 


MENSURATION  OF  SUPERFICIjt:^ 


By  the  Sliding  Rule. 

Set  12  on  D  to  -}  of  the  circumference  on  C,  and  against  the 
length  on  D  is  the  answer  on  C. 

By  Gttnter. 

The  extent  from  12  to  i  of  the  circumference  doubled,  or  twice 
turned  oyer,  will  reach  from  the  length  to  the  content. 

ExAMP.  3.  In  the  frustum  of  a  tiiangular  pyramid, 
the  side  of  the  greater  base  AC;=15,  as  before,  the 
side  of  the  leys  BD=6,  and  the  height  EF=40,  to 
find  the  content. 


15=AC. 
6=:BD. 

9=difference  of  the  sides. 
9 

i)81=:square  of  the  diiference. 

27=i  of  the  square. 


15 

e 

90 
Add  27 


117 
'433  multiplier. 


351 
351 

468 


50-661  =mean  area. 
40=height. 


2026-440=content. 
Or,  if  it  be   a  tapering  three  square  stick  of  timber,  you  may 
find  the  area  midway  from  end  to  end,  then,  as  144  is  to  that  area, 
so  is  the  iength,  taken  in  feet,  to  the  content  in  feet. 

Ex  AMP.  4.     To  ineamre  the  Frustum  of  a  Cone. 

Rule.  Multiply  the  diameters  of  the  two  bases  together,  and 
to  the  product  add  one  third  of  the  square  of  the  difference  of  the 
diameters  :  then  multiplying  this  sum  by  -7854,  it  will  be  the  mean 
area  between  the  two  bases,  which  being  mullipiied  by  the  length 
of  the  frustum,  will  give  the  solid  content. 

Or,  to  the  areas  of  the  top  and  bottom  add  the  square  root  of 
the  product  of  those  areas,  and  the  sum,  multiplied  by  one  third 
of  the  height  of  the  frustum,  will  give  the  solidity. 

When  fjijfures  run  uniformly  taper  ;  but  not  to  a  point  (they  be- 
ing considered  as  portions  of  the  cone  or  pyramid)  we  may  find 
the  solidity  by  supplying  what  is  wanting  to  complete  the  figure, 
and  then  deducting  the  part  (iut  ofi'. 

A  general  rule  for  completing  every  straight  sided   solid,  nshose  ends 
are  parallel  and  similar. 

As  the  difference  of  the  top  and  bottom  diameters  is  to  the  per- 
pendicular height,  (or  depth  which  is  the  same  :)  so  is  the  long^'Si 
diameter  to  the  altitude  of  the  whole  cene  or  pyramid. 


AND  SOLIDS. 


471 


ExAMp.  1.  The  former  cone  in  Art.  32,  Exanip.  3,  being  cut  off 
in  the  middle,  the  greater  diameter  AC  is  13,  the  less  BD  6i,  and 
height  EF  24,  to  fmd  the  content  of  the  frustum. 

AC=13  inches.  '  13 

BD=6-5  inches.  6-5 


65 
78 

84'6 
Add  14083 


98-583 
•7854 


394332 
492915 
788664 
690081 


6  5=difference. 
6-5 

325 
390 

i4-083=i  of  the  square. 

144)1858'248(12-9045  feet  content 
144 


77-427|0882=mean  area. 
24  feet=length. 


418 

288 


309708 
154854 


1302 
1296 


1858-248=content. 


648 
576 


720 
720 
ExAMP.  2.     What  number  of  barrels,  each  32  gallons  of  Ale 
measure,  is  contained  in  a  cistern  whose  largest  diameter  is  6  feet, 
and  smallest  diameter  5  feet,  and  whose  depth  is  8  feet  ? 
6x5=30 

12  i 


6--5I 


:1=    01- 
3  ^J 


30i  mean  diameter. 
•7854 


23-5620 
•26ia 


23-8238  mean  area. 
8 


190-5904  content  in  feet. 

1728  inches  in  a  solid  foot. 


329340  2112  cubic  inches,  which  divided  bj-  9024,  the  cubic 
mches  in  a  barrel  or  32  gallons,  gives  36-5  barrels  nearly,  Ans. 


472  MENSURATION  OF  SUPERFICIES 

If  the  answer  had  been  required  in  Beer  Measure,  where  the 
barrel  contains  36  gallons,  the  answer  would  have  been  32*4  bar- 
rels. 

Note.  If  when  the  end  diameters  of  a  conical  cistern  are  giv- 
en, it  is  required  to  find  the  length  of  the  cistern  to  contain  a  cer- 
tain number  of  barrels;  divide  the  cubic  fieet  contained  in  the 
number  of  barrels  by  the  mean  area,  and  the  quotient  will  be  the 
height. 

Let  the  mean  area  be  as  in  the  last  Ex.  to  find  the  length  of  the 
cistern  to  contain  60  barrels  of  32  gallons  of  Ale  measure. 

26111  111  &c.=cubic  feet  in  50  barrels,  which  divided  by 
i23  8238,  the  mean  area,  gives  1095  feet,  for  the  length  of  the 
cistern,  Ans. 

To  find  the  diameters  of  the  cistern,  when  the  content,  and 
length,  and  difference  of  the  diameters,  are  given,  see  Art.  53. 

Art.  34.     To  measure  a  Sphere  or  Globe, 

Definition.  A  sphere  or  globe  is  a  round  solid  body,  in  the  mid- 
dle of  which  is  a  point,  from  which  all  lines  drawn  to  the  surface 
are  equal. 

Rule.  Multiply  the  cube  of  the  diameter  by  -5236,  and  the 
product  will  be  the  solid  content. 

Or,  multiply  the  circumference  by  the  diameter,  which  will  give 
the  superficial  content ;  then  multiply  the  surface  by  one  sixth  of 
the  diameter,  and  it  will  give  the  solidity. 

Or,  multiply  the  cube  of  the  diameter  by  11,  and  the  product 
divided  by  21,  will  give  the  solidity. 

ExAMP.  The  diameter,  AB,  of  a  globe,  is  4-5  feet  ;  to  find  the 
solid  content. 

4-5 
4-5 

225 
180 

20-26 
4-5 

10125 
8100 

91-125 

•523G 

546750 
273375 
182250 
455G25 


47-7130500 


AND  SOLIDS. 


473 


'  Note.  If  the  circumference,  or  greatest  circle  of  the  sphere, 
be  given,  multiply  the  cube  of  it  by   016887  for  the  content. 

'I'he  surface  of  the  globe  may  be  found  by  multiplying  the 
square  of  the  diameter  by  3' 1416  ;  or  by  multiplying  the  area  of 
its  greatest  circle  by  4,  or  the  square  of  the  circumference  by 
•3183. 

When  the  solidity  of  a  globe  is  given,  the  diameter  maybe  found 
by  dividing  the  solidity  by  -5236,  and  extracting  the  cube  root  of 
the  quotient. 

Or,  if  the  circumference  be  required,  divide  the  solidity  by 
•016887,  and  the  cube  root  of  the  quotient  will  give  it. 

Art.  35,  To  measure  the  Solidity  of  a  Frustum  or  Segment  of  aGlobe^ 
Definition.      The  frustum  of  a  globe  is  any  part  cut  off  by  a 

plane. 

Rule.     To  three  times  the  square  of  the  seraidiaraeter  of  the 

base,  add  the  square  of  the  height;    then  multiplying  that  sum 

by  the  height,  and  the  product  by  -5236,  you  will  have  the  solid 

content. 

ExAMP.     The  height  BL)  beirig   9  inches,   and  the  diameter  of 

the  base  AC  24  inches  :   to  find  the  content. 

12=8emidiam6ter.        4617  jy 

12  -5236 


144=square. 
X    3 


27702 
13851    A 
9234 
23085 


432      .  f. 

Add  9X9=  81=  J  f?"^^.^^, 

—      ^  ^^'^  ^^•S^^-2417-4612=solid  content. 
613 
X     9=height. 


4617 

To  measure  the  Surface  of  a  Frustum  or  Segment  of  a  Globe. 
Rule.  Find  the  diameter  of  the  globe  by  Art.  24,  and  the  sur- 
face of  the  whole  globe,  by  Art.  34  ;  then,  as  the  diameter  of  the 
globe  is  to  the  height  of  the  frustum  ;  so  is  the  surface  of  the 
globe  to  the  surface  of  the  frustum  ;  then,  by  Art.  15,fifjd  the  area 
of  the  base  ;  add  these  two  together,  and  the  sum  will  be  the  whole 
surface  of  the  frustwni. 

Art.  36.     To  measure  the  middle  Zone  of  a  Globe. 

Definition.  This  part  of  a  globe  is  somewhat  like  a  cask,  two 
equal  segments  being  wanting,  one  on  each  side  of  the  axis. 

KuLE.  To  twice  the  square  of  the  middle  diameter,  add  the 
square  of  the  end  diameter;  multiply  that  sum  by  -7854,  and  that 
product,  multiplied  by  one  third  of  the  length,  will  give  the  go* 
lidity. 

M  3 


474 


MENSURATION  OF  SUPERFICIES 


Or,  to  four  times  the  square  of  the  middle  diameter  add  twicr 
the  square  of  the  end  diameter  ;  that  sum  multiplied  by  '7854,  and 
that  product  by  one  sixth  of  the  length,  will  give  the  solidity. 

Note.  This  rule  is  applicable  to  the  frustum  of  a  cone  or  py- 
ramid. 

If  the  middle  diameter  of  a  zone  be  20  inches,  the  end  diame- 
ters each  16  inches,  and  length  12  inches  :  Required  its  solidity  ? 

20x20x2+ 16xl5x-7854x4=3317-6296,  Ans. 

Art.  37.     To  measure  a  Spheroid. 

Definition.  A  spheroid  is  a  solid  body  like  an  egg^  only  both  it§ 
ends  are  the  same. 

RuLEi.     Multiply  the  square  of  the 
diameter  of  the  greatest  circle,  viz.  the 
diameter  of  the  middle  (DB  in  the  fig- 
ure) by  the  length  AC,  and  that  pro-    , 
duct  by  -5236,  and  you  will  have  the  -A, 
solidity. 

ExAMP.  Tke  diameter  BD  being  20, 
and  the  length  AC  30,  to  find  the 
content. 

20x20x30x-6236=6283-2,  Ans. 

Art.    38.   To  measure  the  middle  Frustum  of  the  Spheroid, 

Definition.  This  is  a  cask  like  solid,  wanting  two  equal  seg- 
ments to  complete  the  spheroid. 

Rule.     The  same  as  in  Article  36. 

If  the  middle  and  end  diameters  of  the  middle  frustivm  of  a  sphe- 
roid be  40  and  30  inches,  and  its  length  50 ;  what  is  its  solidity  ? 

50-^-3— 16  6,  then40x40x2l|-30x30X-7854xl6-6=53454-324,Ans. 

Art.  39.     To  measure  a  Segment^  or  Frustum  of  a  Spheroid. 

Definition.  This  is  a  part  of  a  spheroid  made  by  a  plane,  par- 
allel to  its  greatest  circular  diameter. 

Rule.  'To  four  times  the  square  of  the  middle  diameter  add 
the  square  of  the  base  diameter,  then  multiply  that  sum  by  -7854, 
and  the  pro<luct  by  one  sixth  of  the  altitude,  and  it  will  give  the 
solidity. 

If  the  ba^e  diameter  of  the  end  frustum  of  a  spheroid  be  36,  di- 
ameter at  the  middle  of  the  height  30,  and  the  height  20  inches  ; 
required  its  solidity  ? 

30X30X4-f36X36X •7854X3-3=  12689-55-f,  Ans. 

Art,  40.   To  measure  a  Faraholick  Conoid. 

Definition.  This  solid  may  be  generated  by  turning  a  semipara- 
bola  about  its  abscissa  or  altitude. 


AND  SOLIDS,  475 

KuLE.  As  a  jiarabolick  conoid  is  half  of  its  circiMHScribing 
cylinder,  of  the  same  base  and  altitude  ;  multiply  the  area  of  the 
base  by  half  the  height  for  the  solidity. 

If  the  diameter  of  the  base  of  a  parabolick  conoid  be  40  inches, 
and  its  height  42;  what  is  the  solidity  ? 

40X40X-7854X21=26389-44,  Ans. 

Art.  41.     To  measure  the  lower  Frustum  of  a  Parabolick  Conoid. 

Definition.  This  solid  is  made  by  a  plane  passing  through  the 
conoid  parallel  to  its  base. 

KuLE.  Multiply  the  sum  of  the  squares  of  the  diameters  of  the 
bases  by  -7854,  and  that  product  by  half  the  height,  for  the  solidity. 

If  the  diameters  of  a  frustum  of  a  parabolick  conoid  be  40  and 
3Q  inches,  and  its  height  20  inches  ;  required  its  solidity. 

40X40-f30X30X -7854X10=19635,  Ans. 

Art.  42.     To  measure  a  Parabolick  Spindle, 

Definition.  This  solid  is  formed  by  an  obtuse  parabola,  turned 
about  its  greatest  ordinate. 

Rule.  This  solid  being  eight  fifteenths  of  its  ieast  circumscrib- 
ing cylinder,  multiply  the  area  of  its  middle  or  greatest  diameter 
by  eight  fifteenths  of  its  perpendicular  length,  and  it  will  give  its 
solidity. 

If  the  diameter  at  the  middle  of  a  parabolick  spindle  be  20  ioch^ 
es,  and  its  length  60 ;  required  its  solidity. 

20X20X-7854X32  (=60x8—15)  =10053-12,  Ans. 

Art.  43.  To  measure  the  middle  Zone,  or  middle  Frustum^  of  a  Par- 
abolick Spindle. 

Definition.  This  is  a  cask  like  solid,  wanting  two  equal  ends  of 
said  spindle. 

Rule.  To  the  sum  and  half  sum  of  the  squares  of  the  two  di- 
ameters add  three  tenths  of  the  difference  of  their  squares,  which 
multiply  by  a  third  of  the  length,  and  the  product  will  be  the  so- 
lidity. 

If  the  middle  and  end  diameters  of  the  middle  frustum  of  a  par- 
abolick spindle  be  40  and  30  inches,  and  its  length  60 ;  what  is  its 
solidity  ? 

40X40=1600     1600— 900=;=  700  the  difference  of  the  squares. 

30X30=  900       700X'3=2l0=three  tenths  of  do.  then. 


Sum=2500     25004-1250-}-210x20(=iof60)=79200,  Ans. 
Half  sum=1250 

Art.  44.     To  measure  a  Cylindroid,  or  Prismoid. 

Definition.  A  cylindroid  is  a  solid  somewhat  like  tha  frustum  of 
a  cone,  one  base  may  be  an  ellipsis,  and  the  other  a  disproportion^ 
a'l  ellipsis  or  circle. 


476  MENSURATION  OF  SUPERFICIES 

A  prlsnaoid  is  a  solid  somewhat  like  the  frustum  of  a  pyramid, 
but  its  bases  are  disproportional. 

Rule.  The  same  as  for  the  frustum  of  a  cone- or  pyramid  :  or, 
to  the  areas  of  both  bases,  add  a  mean  area,  that  is,  the  square  root 
of  the  product  of  the  two  bases,  then  multiply  that  sum  by  a  third 
of  the  height  or  length,  and  it  will  give  the  solidity. 

If  the  diameters  of  the  greater  base  of  a  cylindroitl  be  30  and  20 
inches,  the  diameter  of  the  less  base  12,  and  length  60  inches  ; 
^hat  is  the  soliditv. 

30X20=:600    "I 

12X12=144         1037-9X-7854X20  (=60—3)==: 
V'144X600=293-9  Y  1630333,  Ans. 

1037-9  J 
If  the  diameters  of  the  greater  base  of  a  prismoid  be  SO  and  2^ 
inches,  the  less  base  20  by  10  inches,  and  length  30  inches:  What 
is  its  solidity  ? 

30X20=600    "" 
20X10=200 

V6M^i^=346-4  ^114^"^^^^  (.=-f  ■^^)=^^^^4«^^i^'^^ 
^  in  mcbes. 

1146-4J 

Note.  To  find  the  solidity  of  a  Wedge,  add  the  length  of  th? 
edge  to  twice  the  length  of  the  base,  and  multiply  the  sum  by  the 
prodiict  of  the  height  of  the  wedge  and  the  breadth  of  the  base, 
and  one  sixth  of  this  product  will  be  the  solidity. 

Let  the  base  of  a  wedge  be  27  by  8,  the  edge  36,  and  the  height 

^^      .        2X27+36X8X42      _,^.^    . 

42  ;  then =5040,  Ang. 

6 

Art.  45.     To  measure  a  Solid  Ring. 

Rule.  Measure  the  internal  diameter  of  the  ring,  and  its  girth, 
or  circumference  :  then  multiply  the  girth  by  3I83I,  and  the  pro- 
duct will  be  the  diameter  of  the  wire,  which  add  to  the  internal  di- 
ameter ;  multiply  this  sum  by  31416,  and  the  product  will  be  the 
length  of  a  cylinder  equal  to  the  ring  of  the  same  base.  Then 
the  area  of  a  section  of  the  ring  multiplied  by  the  length  of  the 
said  cylinder  will  give  the  solidity  of  the  ring. 

If  an  iron  ring  be  12  inches  in  girth,  and  its  internal  diameter  be 
20  inches  ;   what  is  its  solidity  ?  ______ 

•31831  Xl2=3'8=ring's   diameter.      20-|-3'8X31416=74  77  the 
kngth  of  a  cylinder  equal  to  the  ring  :   And 

3  8X3  8X-7864X74-77=847-97=solidity. 

Art-  46.     To  medmre  the  Solidity  of  any  irregular  Body^  whose  di- 
inensions  cannot  be  taken. 

Take  any  regular  vessel,  either  square  or  round,  and  put  the 
irregular  body  into  it :  pour  so  much  water  into  the  vessel  as  will 
exactly  cover  the  bodj,  and  measure  the  dry  part  from  the  top  of 


AND  SOLIDS. 


477 


the  vessel  to  the  water,  then  take  out  the  body,  and  measure 
again  from  the  top  of  the  vessel  to  the  water,  and  subtract  the  first 
measure  from  the  second,  and  the  difference  is  the  fall  of  the  water  ; 
then,  if  the  vessel  be  square,  multiply  the  side  by  itself,  and  that  pro- 
duct by  the  fall  of  the  water,  and  you  will  have  the  content  of  the 
body  ;  but  if  it  be  a  long  square,  multiply  the  length  by  the  breatith, 
and  that  product  by  the  fall  of  the  water  ;  or,  lastly,  if  it  be  a  round 
vessel,  multiply  the  square  of  the  diameter  by  '7854,  and  that  pro- 
duct by  the  lall  of  the  water,  and  you  will  have  the  content. 


ExAMP.   1.     A  body" 
being  put  into  a  vessel 
18    inches    square,  on 
taking  out   the   body, 
the  water  sunk  9  inch- 
es ;  required  the  con- 
tent of  the  body  ? 
18  inch.  =  1-5  foot. 
9  inch. =  75  foot. 
1-6    X    1  5  X    '75  = 
1-6875  foot,  content. 


Exam?.  2.  A  body 
put  into  a  cistern  4 
feet  by  3,  on  tak- 
ing it  out,  the  wa- 
ter fell  6  inches ; 
required  the  con-^ 
tent  of  the  body  ■? 


4  X  3  X  -5=6  feet,    body  ? 


ExAMP.  3.  A  body 
being  put  into  a  round 
tub,  whose  diameter 
was  1-5  foot,  on  taking 
out  the  body,  the  wa- 
ter fell  1-5  foot;  what 
1  was  the  content  of  the 


content. 


15Xl'5X-7S54Xl'5: 
2-65  feet,  content. 


I 


Of  the  five  Regular  Bodies. 

There  are  five  solids  contained  under  equal  regular  sides,  which 
by  way  of  distinction,  are  called  the  Jive  regular  bodies. 

'J'he«e  are  the  Tetraedron,  Ihe  Hexaedron  or  Cube,  ihe  Octaedron^ 
the  Dodecaedron,  and  the  Eicosiedron.  The  measuring  of  the  cube 
was  shewn  at  Art.  28.  I  shall  n-ow  show  how  to  measure  the  other 
four  by  the  following  Table,  which  is  the  shortest  method. 

A  Table  of  the  solid  anct  superficial  content  of  each  of  the  Jive  bodies, 
the  sides  being  unity,  or  1. 


Names  of  the  .Bodies. 

Soiidity. 

Superficies. 

1  fctraedron. 

Hexaedron. 

Octaedron. 

Eicosiedron. 

Dodecaedron. 

011785 
1- 

0-4714 

2181695 
7-663119 

1-73205 
6- 
3  464 

8-66025 
20-6457 

All  like  solid  bodies  being  in  proportion  to  one  another  as  the 
cubes  of  their  like  sides,  the  solid  content  of  any  of  these  bodies 
may  be  found  by  n^ulliplying  the  cubes  of  their  sides  by  the  npm- 
bers  in  the  second  column  under  Solidity  ;  and  their  superficies,  by 
multiplying  the  squares  of  their  sides  into  the  numbers  in  the  third 
column  under  Superficies. 

Of  the  Tetraedron. 
This  solid  is  contained  under  four  equal  and  equilateral  triangles, 
that  is,  it  is  a  triangular  pyramid  of  ibur  equal  faces,  the  side  of 
whose  base  is  equal  to  the  slant  height  of  the  pyramid,  from  the 
angles  to  the  vertex. 


178 


MENSURATiON  OF  SUPERFICIES 


Art.  47.     The  side  of  the  Tefraedron  being  3,  to  find  the  solid  and 
superficial  content. 

Cube=3x3x3=27,and  27x- H  785=3  18195=solidity. 
Square==3x3=»9,  and  9x1 '73205=1 568845=snpcrlicies. 

Of  the  Octaedron. 
This  solid  is  contained  under  eight  equal  and  equilateral  triangles, 
which  may  be  conceived  to  consist  of  two  quadrangular  pyramids 
of  equal  bases  joined  together,  the  sides  of  whose  bases  are  equal 
io  the  given  sides  of  the  triangles,  under  which  it  is  contained. 

Art.  48.     The  side  of  an  Octaedron  being  3,  to  find  the  solid  and  su^ 
perficial  content. 
Cube=3x3x3=27,  and  27x4714=12  7278=solidity. 
Square=3x3=9,  and  9x3  464=3M76=superacies. 

Of  the  Dodecaedron. 
This  solid  is  contained  under  12  equilateral  pentagons,  and  may 
be  conceived  to  consist  of  twelve  pentagonal  pyramids,  of  equal 
bases  and  altitude,  whose  vertices  meet  in  the  centre  of  the  dode- 
Cetcdron. 

Art.  49.     The  side  of  a  Dodecaedron  being  3,  to  find  the  solid  ani 
superficial  content. 
Cube=3x3x3=27,  and  27x7-663119=206'904. 
Square=3x3=9,  and  9x20-6457=185-81 13. 

Of  the  Eicosiedron. 
This  solid  is  contained  under  twenty  equal  and  equilateral  trian- 
gles, and  may  be  conceived  to  consist  of  twenty  equal  triangular 
j^yrainids,  whose  vertices  all  meet  in  the  centre. 


Art.  50.      The  side  of  an  Eicosiedron  being  S^to  find  the  solid  and 
superficial  content. 

Cube=3x 3X3=27,  and  27x2-18169=58-90563=solidily. 

Square=3X3=9,  and  9x8'66025=77-94225=superficies. 

As  the  figures  of  some  of  these  bodies  would  give  but  a  confused 
idea  of  them,  I  have  omitted  them  ;  but  the  following  figures,  cut 
out  in  pasteboard,  and  the  lines  cut  half  through,  will  fold  up  into 
the  several  bodies. 


Tetraedron. 


Ilexaedron, 


Octaedron, 


u^^ 


AND  SOLIDS.  47.1 

Dodecaedron.  Ekosiedron. 


OF  CASK  GAUGING. 

Among  the  many  different  canons  drawn  from  Stereometry,  for 
Gauging  casks,  the  following  is  as  exact  as  any. 

Take  the  dimensions  of  the  cask  in  inches,  viz.  the  diameter  at 
the  bung  and  head,  and  length  of  the  cask;  subtrfict  the  head  di- 
ameter from  the  bung  diameter,  and  note  the  dift'erence. 

If  the  staves  of  the  cask  be  ranch  curved  or  bulging  between  the 
bung  and  the  head,  multiply  the  difference  by  -7;  if  not  quite  so 
curve,  by  -65;  if  they  bulge  yet  less,  by  -6;  and  if  they  are  almost 
or  quite  straight,  by  '55,  and  add  the  product  to  the  head  diameter  ; 
the  sura  will  be  a  mean  diameter,  by  which  the  cask  is  reduced  to 
a  cylinder. 

Square  the  mean  diameter,  thus  found,  then  multiply  it  by  the 
length  ;  divide  the  product  by  359  for  ale  or  beer  gallons,  and  by 
294  for  wine  gallons. 

Note  1.  The  length  is  most  conveniently  taken  by  callipers,  al- 
lowing, for  the  thickness  of  both  heads,  1  inch,  \\  inch,  or  2  inch- 
es, according  to  the  size  of  the  cask;  but  if  you  have  no  callipers, 
do  thus  ;  measure  the  lerrgth  of  the  stave,  then  take  the  depth  of 
the  chimes,  which  with  the  thickness  of  the  head,  being  subtracted 
from  the  length  of  the  stave,  leaves  the  length  within. 

Note  2?  You  must  take  the  head  diameter,  close  to  its  outside, 
and,  for  small  casks,  add  three  tenths  of  an  inch  :  for  casks  of  30, 
40,  or  50  gallons,  4  tenths,  and  for  larger  casks,  5  or  6  tenths,  and 
the  sum  will  be  very  nearly  the  head  diameter  within.  In  taking 
the  bung  diameter,  observe,  by  moving  the  rod  backward  and  for- 
ward, whether  the  stave,  opposite  the  bung,  be  tliicker  or  thinner 
tjjan  the  rest,  and  ifitbe,  make  allowance  accordingly. 

By  the  Sliding  Rule. 

On  D  is  18'91,  the  gauge  point  for  ale  or  beer  gallons,  marked 
AG,  and  17  14,  the  gauge  point  for  wine  gallons,  marked  WG  ; 
set  the  gauge  point  to  the  length  of  th^  cask  on  C,  and  against  the 
mean  diameter,  on  D,  you  will  have  the  answer  in  ale  or  wine  gal- 
Ions  accordingly  as  which  gauge  point  you  make  use  of. 

By  the  Scale. 

Take  the  extent  from  the  guage  point  to  the  mean  diameter,  set 
one  loot  of  the  dividers  in  the  length,  and  tnrnina-  Ihcm  twice  over, 
they  will  point  o'jt  th^^  content 


480  MENSURATION  OF  SUPERFICIES 

Art.  ol.  Required  the  content  in  ale  and  wine  gallons,  of  ii 
cask,  whose  bung  diameter  is  35  inches,  head  diameter  27  inches, 
and  length  45  inches  ? 

Bung  diarneter=35     Square  of  the  diameter==1062'76 

Length=  45 


Head  diamet< 

5r=27 

Difference 

=  8 

•7 

Add  the  head  dia.: 

66 

=27 

Mean  diameter 

=32  6 
326 

1966 
652 
978 

631380 
425104 


359)47824-20(133-21 

[ale  galL 

294)47824-2(l62-66  wine  gall. 


Squared  1062-76 

Art.  52.  A  round  mash  tub  is  42  inches  diameter  at  the  top, 
within,  and  36  inches  at  the  bottom,  and  the  perpendicular  height 
48  inches ;  required  the  content  in  beer  and  wine  gallons  ? 

This  being  the  lower  frustum  of  a  cone,  to  the  product  of  the  di- 
ameters add  ^  of  the  square  of  their  difference  ;  multiply  this  sum 
by  the  length,  and  it  will  give  the  solidity  in  such  parts  as  the  di- 
mensions are  taken  in.  l{  they  be  taken  in  inches,  divide  by 
359  for  beer,  and  294  for  wine  galloris. 

42^36  ,ii_±iil!zi^X48-:-  5  359=2031  ale  gallons. 
4^x  jb-t-  ^  X  !«—  ^  294=248|  wine  gallons. 

Art.  63.  Let  the  difference  of  diameters  of  this  tut)  be  6  inch- 
es, the  heio;ht  48  inches,  and  the  content  203f  gallons,  to  find  the 
diameters  ? 

xAlnltiply  the  content,  if  beer  measure,  by  359  ;  if  vvine  measure, 
by  294,  and  divide  the  product  by  the  length  :  from  the  quotient 
subtract  i  of  the  square  of  the  difference  of  the  diameters;  to  this 
remainder  add  the  square  ofi  the  difference  of  the  diameters,  and 
extract  tiie  square  root  of  the  sum  ;  from  the  square  root  subtract  } 
the  difference  of  the  diameters,  and  it  will  giv«  the  least  diameter 
to  great  exactness,  to  which  add  the  difference  of  the  diameters, 
and  the  sum  is  the  greatest  diameter. 


203-75X359        6X6 

v/ [-3x3—3=36,  and  36-f  6=42. 

48  3 

The  diameters  are  36  and  42. 

Tlie  content  of  any  veirsel  in  gallons,  ^c.  may  be  thus  found  : 
measure  the  inside  of  the  vessel,  according  to  the  rule  of  the 
figure,  and  find  the  content  in  cubick  inches  j  then, 


AND  SOLIDS.  481 


fl7. 

Divide  by  <    ^c^ 

(215 


728         \  ^cubickfeet. 

50-425)  (bushets. 


282         f  and  the  quotient  will  1  ale  orbeer  gallons. 
231         t     be  the  content  in     J  wine  gallons. 


Art.  54.     To  ullage  a  Cask,  lying  on  one  side,  by  the  Gauging  Rod, 
when  the  Bung  Diameter,  and  the  Content,  out,  or  both  are  greater 
or  less  than  the  Table  on  the  Rod  is  made  for. 
Rule.     As  the  bung  diameter  of  the  cask  to  be  measured,  is  to 
the  bung  diameter  that  the  table  is  made  for ;  so  are  the  dry  inch- 
es of  the  cask,  to  a  fourth  number,  which  find  in  the  table  on  the 
rod,  and  note  the  number  of  gallons  answering  to  it.     Then  as  the 
content  of  the  ca.'^k  that  the  table  is  made  for,  is  to  the  content  of 
the  cask  to  be  measured  ;  so  is  the  number  of  gallons  answering 
to  the  aforesaid  fourth  number,  to  the  number  of  gallons  your  cask 
wants  of  being  full. 

Art.  55.     3b  find  a  Ship's  Burthen,  or  to  Gauge  a  Skip. 

There  is  such  a  diversity  in  the  forms  of  ships,  that  no  general 
rule  can  be  applied  to  answer  all  varieties  ;  however,  the  follow^ 
ing  rules  are  practised. 

Rule  1.  Multiply  the  breadth  at  the  main  beam,  half  the  breadth, 
and  length  together  ;  divide  the  product  by  94,  and  the  quotient  is 
the  tons. 

Rule  2.  Divide  the  continued  product  o{  the  length,  breadth, 
and  depth,  in  feet,  by  100,  for  ships  of  war,  and  95  for  merchant 
ships,  in  which  nothing  is  allowed  for  guns,  &c.  and  the  quotient 
is  the  tons. 

Rule  3.  Take  the  length  from  the  stern  post  to  the  upper  part 
of  the  stem  ;  subtract  two  thirds  of  her  breadth  from  that  length  ; 
multiply  the  remainder  by  the  whole  breadth,  and  that  product  by 
half  the  b/eadth,  in  feet,  and  divide  by  100  for  war,  and  94  for 
merchant  tonnage. 

Rule  4.  The  weight  of  a  ship's  burthen  is  half  the  weight  of 
water  she  can  hold. 

What  is  the  tonnage  of  a  ship,  whose  length  is  97  feet,  breadth 
31  feet,  and  depth  15ifeet. 

By  Rule}  St.        '  By  Rule  2d, 

\  breadth  15-5  Length  97  . 

Breadth    31  Breadth  31 

165  97 

465  291 


480  5  3007 

Length       97  Depth=  155 


33635  15035 

43245  15035 


3007 


!l4)466p8-5(495  83  tone.  O5Vl6608-5(490-61  tons. 

Carried  ever,  '   Cfrrierl  ovor. 

N  3 


182  MENSURATION  OF  SUPERFICIES  &c. 


Brought  over. 

Brought  over. 

94)46608-6(49583  tons. 

95)46608-5(490-61  tons. 

376 

380 

•       100)46608-5(466  tons. 

900 

400 

860 

846 

855 

660 

548 

600 

585 

470 

570 

608 

785 

600 

150 

752 

85 

95 

330 

55 

282 

48 

By  Rule  M. 
Length=97 

Subtract  \ 

ofbreadth=20-66 

76-33 

Multiply  by  the  breadth      31 

7633 

22899 

2366-23 

Multiply  by  1  breadth        15-5 

1183115 

1183115 

236623 

94)36676-565(390'176  tans. 


Allowing  the  Cubit,  as  it  is  found  by  modern  travellers,  to  he  22  inches 
the  content  of  Noah'' s  Ark  is  as  follows,  viz. 

Cubits. 
Length  of  the  keel,  300^  Its  burthen  as  a  man  of  war 

Breadth  by  the  midship  beam,    50  >      27729  tons. 
Depth  in  the  hold,  30)  As  a  merchant  ship,  29188*6  ts 


QUESTIONS  IN  MENSURATION.  4?; 


(QUESTIONS  IN  MENSURATION. 

1.  THE  largfest  of  the  Egyptian  pyramids  is  square  at  the  base, 
and  m  easures  693  feet  on  a  side  :  how  much  ground  does  it  cover  ? 

696  X393  1764 

•-^=^—  =  1764  poles,  and  7^=11  acres  and  4  poles,  Ans. 

2.  What  difference  is  there  between  a  floor  20  feet  square,  and 
two  others,  each  10  feet  square  ? 

20X20—10X10+  10X10=200  feet,  Ans. 

3.  There  is  a  square  of  2500  yards  in  area  :  what  is  each  side 
of  the  square,  and  the  breadth  of  a  walk  along  one  side  and  one 
end,  which  may  take  up  just  one  half  of  the  square  ? 

2500 

V2500=50  yards,  each  side.     V"^— ^35-35,   and   50— 33-35 

=:14'65  yard?,  breadth  of  the  walk,  Ans. 

4.  A  pine  plank  is  16  feet  and  5  inches  long,  and  I  would  have 
just  a  square  yard  slit  off:  at  what  distance  from  the  edge  must 
the  line  bo  drawn  ? 

A  square  yard=1296  inches,  and  16  feet  5  inches=197  inches. 

1 296 
Therefore,  -r^=6-J-f^  inches,  Ans. 

5.  If  the  area  of  a  triangle  be  ,900  yards,  and  the  perpendicular 
40  yards:  required  the  length  of  the  base  ? 

900X2 

■       ■     =45  yards,  Ans. 

6.  If  the  three  sides  of  a  plane  triangle  be  24,  16,  and  12  perch- 
es :  required  its  area  ? 

24+16+12 

2 =2^5  26—24—2;  26—16=^10;    26—12=14,   and 


2 


V26X  14X10X2=85-32  perches, =area.     Again,  as  24:  16+12  :: 
16  — 12  :  4-6+,  the  difference  of  the  segments  of  the  base  ;  then, 

4-6+  ' 

12—  ~--=9'6,  and  >/ 12X12— 9-6x9-6=7-ll  the  perpendicular 

on  the  longestside;  whence  24^2X7- 11  =85*32,  area  as  above. 

7.  Required  the  area  of  a  circular  garden,  whose  diameter  is  12 
rods?  12X12X'7854  =  113  0976  poles,  Ans. 

8.  The  wheel  of  a  perambulator  turns  just  once  and  a  half  in 
a  rod  :  what  is  its  diameter  ? 

16  5X|  =  ll  circumference,  and  l]X-31831=3i  feet,  Ans. 

9.  Agreed  for  a  platform  to  the  curb  of  a  round  well,dt7id.  per 
cquare  foot  :  the  inward  part,  round  the  mouth  of  the  well,  is  36 
inches  diameter,  and  the  l>reail(h  of  (he  platform  was  to  be  15| 
inches  :   wiiat  will  it  como  (o  .' 


484  QUESTIONS  IN  MENSURATION. 


36-|-l5-5x2=67the  greatest  cIiam.;67x67X-7864—.36x36X'7854 
2507-8722 

s=  ~ — =17'4157  square  feet,  at  7|d.  per  foot, =10s.  10,\(1. 

[Aug. 

10.  Required  the  difterence  between  the  area  of  a  circle,  whose 
radius  (or  semidiameter)  is  60  yards,  and  its  greatest  inscribed 
square  ? 

60X2=100  the  diameter,  and  lOOXlOOX-7854=:7864  the  area 
of  the  circle  ;  then,  60X50X2=:6000  the  area  of  the  greatest  in- 
scribed square,  and  7864 — 5000=2854,  Ans. 

11.  There  is  a  section  of  a  tree  25  inches  over  ;  I  demand  the 
difference  of  the  areas  of  the  inscribed  and  circumscribed  squares, 
and  how  far  they  differ  from  the  area  of  the  section  ?  ______ 

26X25— 12  5xT2-5X2=312'5  the  difference  of  the  squares.  25X25 
— 26X26X-7854=134125  the  circumscribed  square,  more  than  the 
section,  and  25X25X-7854~"12-5X  1^-6 X 2=178-375  inscribed 
square,  less  than  the  area  of  the  section. 

12.  Four  men  bought  a  grindstone  of  60  inches  diameter  :  how 
much  of  its  diameter  must  each  grind  off,  to  have  an  equal  share  of 
the  stone,  if  one  first  grind  his  share,  and  then  another,  till  the  stone 
is  ground  awaj,  making  no  allowance  for  the  eye  ? 

Rule.  Divide  the  square  of  the  diameter  by  the  number  of  men, 
subtract  the  quotient  from  tho  square,  and  extract  the  square  rgot 
of  the  remainder,  which  is  th«i  lengtii  of  the  diameter  after  the  first 
man  has  ground  his  share  ;  this  work  being  repeated  by  subtract- 
ing the  same  quotient  from  the  remainder,  for  every  man,  to  the 
last ;  extract  the  square  root  of  the  remainders,  and  subtract  those 
roots  from  the  diameters,  one  after  another  ;  the  several  remain- 
ders will  be  the  answers. 

60  -  From  60 

60  Take  61-9615 


4)3G00  Remains    8  0385— 1st  share 

Quot.=900  From  519615 

Take  424264 
From  3600  


Take     900  Rem.     9  6361~=2d  share. 

v'2700=551-9615,  to  be  taken  from  60. 
Subt.     900  From  42  4264 

V'1800=42-4264,  from  51-0616.       Take  30' 
Subt.     900  — ' 


^/900=30,  from  42-4264.  Rem.   12  4264=3d  share. 

And  30  inches=1th  share. 


QUESTIONS  IN  MENSURATION.  485 

13.  If  a  cubick  foot  of  iron  were  hammered,  or  drawn,  inlo  a 
square  bar,  an  inch  about,  that  is,  ^  of  an  inch  square  :  required 
its  length,  supposing  there  is  no  waste  of  metal  ? 

12X12X12 

•2qX25X4~''*^^^  inches,=576  feet,  Ans. 

14.  Required  the  axis  of  a  globe,  whose  solidity  may  be  just 
equal  to  the  area  of  its  surface  ?  -7854X4 

'5^36  '^^^  inches,  Ans. 
■    15.  A  joist  is  71  inches  wide,  and  2i  thick  ;  but  I  want  one  just 
twice  as  large,  which  shall  be  3f  inches  thick  :  what  will  be  the 
breadth  ?  75X2  25x2 

—— =9  inches,  Ans. 

16.  I  have  a  square  stick  of  timber  18  inches  by  14  ;  but  one 
of  a  third  part  of  the  timber  in  it,  provided  it  be  8  inches  deep, 
will  serve  :  how  wide  will  it  be  ?     18X14 

— - — -~-8=]0i  inches,  Ans. 

17.  A  had  a  beam  of  oak  timber,  18  inches  square  throughout, 
and  25  feet  long,  which  he  bartered  with  B,  for  an  equilateral  tri- 
angular beam  of  the  same  length,  each  side  24  inches  :  required 
the  balance  at  Is.  4d.  per  foot? 

18x18x25 

— —  =56-95,  solidity  of  the  square  beam. 

The  perpendicular  let  fall  on  one  of  the  sides  of  the  triangu- 
lar beam  is  20*7846  inches,  and  the  half  perp.  =  10-3923 ;   then 
■10-3923x24 
— —jj- =1732  foot,  area  at  the  end,  and  1-732x25=43  3 

feet,  solidity  of  the  triangular  beam  ;  therefore  56-25--43-3=12-95 
feet,  at  U.  4d.  per  foot=17s.  3-2d.  balance  due  to  A,  Ans. 

18.  What  is  the  difference  between  a  solid  half  foot,  and  half  a 
foot  solid  ? 

12x12x6 
g^g^^  =4,  therefore,  one  is  but  -|  of  the  other. 

19.  A  lent  B  a  solid  stack  of  hay,  measuring  20  feet  every  way  ; 
sometime  afterward,  B  returned  a  quantity  measuring  every  way 
10  feet :  vy^hat  proportion  of  the  hay  remains  due  ? 

20X20X20—10X10X10=7000  feet=l,  Ans. 

20.  A  ship's  hold  is  75^  feet  long,  18J  wide,  and  7A  deep  :  how 
many  bales  of  goods  3i  ieet  long,  2i  deep,  and  2f  wide,  may  be 
slowed  therein,  leaving  a  gang  way  the  whole  length,  of  3i  feet 
wide  ? 

75-5X18-5'x7-25— 75-5X7-25X3-2"5      on-  ..   ,    , 

~ =38i5-44  bales,  Ans. 

3-5X2-25X2-75 

21.  If  a  stick  of  timber  be  20-1  feet  long,  16  inches  broad,  and 
8  inches  thick,  and  3i  solid  feet  be  sawed  off  one  end :  how  long 
will  the  stick  then  be  ? 

1728x3o 
20-5  —  '^xlT'^'^^  ^"^^^  ^v  inches,  Ans, 


4ati  QUESTIONS  IN  MENSURATION. 

22.  The  solid  content  of  a  square  stone  is  found  to  be  1361  feet  j 
its  length  is  9^  feet  :  what  is  the  area  of  one  end  ?  and  if  the 
breadth  be  3  feet  11  inches,  what  is  the  depth  ? 

136-5X1728  2069-0526 

9-5X12     —^^^'^  20G9C52G  inches,  and   ■;pj =  44022 

inches,  Ans. 

23.  I  wotild  hnve  a  cnbick  box  made  capable  of  receiving  just 
50  bushels,  the  bushel  containing  2150-425  solid  inches  :  what  will 
be  the  length  of  the  side  ?  3 

V2150-4X56=47'55  inches. 

24.  A  statute  bushel  is  (0  be  made  8  inches  high,  and  18i  inches 
diameter,  to  contain  2176  cubick  inches  ;  (though  the  content  of 
the  dimensions  is  but  2160  425  inches)  1  demand  what  the  diame- 
ter of  the  bushel  must  be,  the  height  being  8  inches;  and  what 
the  height,  the  diameter  being  I81  inches,  to  contain  2176  cubick 
inches  ? 

Solidity.  

Height— 8)2176  and  ^/272X1  •273=186  diameter.      18-6X18-5 

X  7854=268  80315=area,  and  the  solidity 

Area=  272         2176-i-2.68-8=8-0956  inches,  height. 

25.  There  is  a  garden  rolling  stone  66  inches  in  circumference, 
and  31  cubick  feet  are  to  be  cut  off  from  one  end,  perpendicular 
to  the  axis  :  where  must  the  section  be  made  ? 

1728x3-5 

A  — ,  10  r — =14-65  inches  from  one  end,  Ans. 

Area=     4125  ' 

26.  I  would  have  a  syringe  of  ]i  inch  diameter  in  the  bore,  to 
bold  a  quart,  wine  measure  :  what  must  be  the  length  of  the  pis- 
ton, sufficient  to  make  an  injection  with  ? 

1  5X1'5X-7854=1-76715,  and  231 --4=57  75  the  cubick  inches 
57-75 
in  a  quart,  then  j.„g^j;=32  679  inches,  Ans. 

27.  If  a  round  pillar,  9  inches  diameter,  contain  5  feet:  of  what 
diameter  is  that  column,  of  equal  length,  which  measures  10  times 
as  much  ? 

As  5  :  9X9  ::  5X10  :  810,  and  -v/810=28-46  inches,  Ans. 

28  There  is  a  square  pyramid,  each  side  of  whose  base  is  30 
inches,  and  whose  perpendicular  height  is  120  inches,  to  be  divid- 
ed by  sections  parallel  to  its  base  into  3  equal  parts  :  required  the 
perpendicular  height  of  each  part? 

30X30X40=36000  the  solidity  in  inches,  now  |  thereof  is  24000, 
3»n(l  1  is  12000.     Therefore, 

^;.3e000:,.0X,20X,.0::jf^O^°J;'SoS'r'>- 

Vi  152000=104-8  Also,  V'576000=83-2.  Then,  120— 104-8 
^152  length  of  the  thickest  part,  and  104-8— 83  2=21-6  length 
of  the  middle  part,  consequently  83  2  is  the  length  of  the  top  part. 

29.  Suppose  the  diameter  of  the  base  of  a  conical  ingot  of  gold 
la  be  3  inches,  and  its  height  9  inches  ;  what  length  of  wire  may 


QUESTIONS  IN  MENSURATION.  487 

be  expected  from  it,  without  loss  of  metal,  the  diameter  of  the  wire 
being  one  hundredth  part  of  an  inch  ? 

3x3x-7854X3=:21-2038  the  solidity  of  the  cone. 
21-2058 

•01  X  Ol'x'^7S^~^^^^^^  inch.  =4  miles,  and  460  yards,  Ans. 

30.  Suppose  a  pole  to  stand  on  a  horizontal  plane  75  feet  in 
height  above  the  surface  :  at  what  height  from  the  ground  must  it 
be  cut  ofl',  so  as  that  the  top  of  it  may  fall  on  a  point  55  feet  from 
the  bottom  of  the  pole,  the  end,  where  it  was  cut  off,  resting  on 
the  stump,  or  upright  part  ? 

As  the  whole  length  of  the  pole  is  equal  to  the  sum  of  the  hy- 
potenuse and  perpendicular  of  a  triangle,  (the  55  feet  on  the 
ground  being  the  base)  this,  as  well  as  the  following  question,  may 
be  solved  by  this 

Rule.  From  the  square  of  the  length  of  the  pole  (that  is,  of 
the  sum  of  the  hypotenuse  and  perpendicular)  take  the  square  of 
the  base  ;  divide  the  remainder  by  twice  the  length  of  the  pole, 
and  the  quotient  will  be  the  perpendicular,  or  height  at  which  it 
must  be  cut  off. 

75X2 

31.  Suppose  a  ship  sails  from  latitude  43°,  north,  between  north 
and  east,  till  her  departure  from  the  meridian  be  45  leagues,  and 
the  sum  of  her  distance  and  difference  of  latitude  to  be  1 35  leagues  : 
I  demand  her  distance  sailed,  and  latitude  come  to? 

135X135— 45X4*5     ^^   ,  .    ^„^„      ,„„       .,         ^    , 
=60  leagues,  and  60X3=180  miles=3  de- 

135X2 
grees  the  difference  of  latitude,  135 — 60=75  leagues  the  distance. 
Now  as  the  vessel  is  sailing  from  the  equator,  and  consequently 
Ihe  latitude  is  increasing  :  Therefore, 

To  the  latitude  saijed  from  43'',00'  N. 

Add  the  difference  of  latitude  3  ,00 

And  the  sum  is  the  latitude  come  to=46  ,00  N; 


BOOK  KEEPING 


BOOK  KEEPING  is  a  systematic  record  of  mercantile  trans- 
actions. 

Every  mercantile  transaction  consists  in  giving  one  thing  for 
another.  This  change  of  property  requires  a  distinct  record  in 
the  books  prepared  for  the  purpose,  so  as  to  enable  the  man  of 
business  to  know  the  true  state  of  his  affairs,  and  of  his  accounts 
with  an  individual. 

The  importance  of  a  correct  knowledge  of  Book  keeping,  to  the 
man  of  business  is  obvious.  His  books  should  exhibit  the  result 
of  each  transaction,  and  the  general  result  of  the  whole. 

Book  keeping  may  be  performed  either  by  Single  or  Double 
Entry. 

The  method  of  book  keeping  by  single  entry  is  the  most  simple, 
and  is  sufficient  for  the  generality  of  Mechanics,  Farmers,  Retail 
Merchants,  &c.  The  method  by  Double  Entry  is  more  systematic 
in  its  principles,  and  more  certain  in  its  conclusions,  and  is  much  to 
be  preferred  for  wholesale  or  any  extensive  business. 

in  Single  Entry  only,  persons  are  entered  as  debtor  and  creditor. 

BOOK  KEEPING  BY  SINGLE  ENTRY. 

In  the  practice  of  single  entry,  two  principal  books,  the  Day 
Book  or  Waste  Book,  and  the  Leger,  and  one  auxiliary  book,  the 
Cash  Book,  are  necessary. 

].  THE  DAY  BOOK  OR  WASTE  BOOK. 

The  Day  Book  sliould  begin  with  an  account  of  all  the  property  ^ 
debts,  &c.  of  the  person,  and  be  followed  by  a  distinct  record  o* 
all  the  transactions  of  the  trade  in  the  order  of  time  in  which 
they  occur. 

Some  accountants  use  also  a  Blotter,  in  which  tlie  changes  of 
property  are  recorded,  and  the  Day  Book  is  only  a  copy  of  the 
Blotter,  written  in  a  more- fair  and   plain  manner 

Each  page  of  the  Day  Book  should  be  ruled  with  three  columns 
on  the  right  side  for  pounds,  shillings,  and  pence,  or  with  two  co- 
lumns for  dollars  and  cents,  as  the  accounts  are  to  be  kept  in  one 
or  the  other  of  these  denominations  of  money. 

The  following  is  the  order  observed  in  making  an  account  in  the 
Day  Book  :  First,  the  date  ;  next,  the  name  of  the  person  with  the 
abbreviation  Dr.  or  Cr.  at  the  right  hand,  as  he  is  debtor  or  credi- 
tor by  the  transaction;  and  then,  the  article  or  articles  with  the 
price  annexed,  unless  the  article  be  money,  and  the  value  carried 


BOOK  KEEPING,  &c.  489 

out  in  the  ruled  columns,  with  the  sum  of  the  whole  placed  direct- 
ly under,  when  there  is  more  than  one  charged.  Thus,  for  ex- 
ample, 


January  1st,  1820. 
David  Bradley, 
To  2  yards  of  broadcloth  at  42s.  per 
To  lOlbs.  loaf  sugar  at  2s.  2d. 

Dr. 

yard, 

£ 
4 

1 
5 

a. 
4 
1 
5 

0 
8 

8 

January  2d. 

Simon  Jones, 
By  3  bushels  of  wheat  at  9s.  Cd. 

Cr. 

I 

8 

6 

The  following  rule  shows  whether  a  person  is  to  be  entered  as 
Dr.  orCr.  on  the  Day  Book.  The  person  who  receives  any  thing 
from  me  is  Dr.  to  me,  and  the  person  from  whom  I  receive  is  Cr. 

Or,  The  person,  who  becomes  indebted  to  me,  whether  by  re- 
ceiving goods  or  money  or  by  my  paying  his  debts,  must  be  enter- 
ed Dr.  ;  and  the  person  to  whom  I  become  indebted,  whether  by 
receiving  from  him  goods  or  money,  or  by  the  payment  of  my 
debts,  must  be  entered  Cr. 

The  following  general  direction  is  to  be  observed  in  keeping  the 
Day  Book,  viz. 

Enter  on  the  Day  Book  every  case  of  debt  or  credit  relating  to 
the  business  in  the  order  of  time  in  which  it  takes  place,  and  in 
language  so  explicit  as  not  to  be  mistaken. 

This  rule  is  most  important,  because  the  Day  Book  is  the  deci- 
sive book  of  reference  in  case  of  any  supposed  mistake  or  error 
in  the  accounts  in  the  Leger. 

2.    THE  LEGER. 

The  various  accounts  of  each  person  are  collected  from  the  Day 
Book,  and  placed  or  posted  under  his  name,  and  on  two  opposite 
pages  of  the  Leger,  as  they  are  Dr.  or  Cr.  The  name  of  the  person 
is  to  be  written  in  large  and  fair  characters  as  a  title,  and  the  ac- 
counts in  which  he  is  Dr.  are  to  be  written  on  the  left  hand  page, 
and  those  in  which  he  is  Cr.  on  the  right  hand  page  of  the  same 
folio.  If  the  name  be  written  only  on  the  Dr.  page,  the  title  of 
the  other  page  is  to  be,  Contra  or  Ca.  Cr.  The  Leger  is  ruled 
with  a  margin  for  the  date  of  each  transaction,  or  with  a  column 
for  the  page  of  the  Day  Book  which  contains  the  account,  or  with 
both.  It  must  also  be  ruled  with  two  or  three  columns  on  the  right 
of  each  page  for  the  denominations  of  money,  as  they  may  be  Dolls, 
and  Cents,  or£  s.  and  d.  If  the  Leger  be  a  wide  Folio,  the  ac- 
counts of  Dr.  and  Cr,  may  be  placed  on  tho  same  page,,  as  in  the 
following  example^ 

0  " 


490 


BOOK  KEEPING 


Dr. 


Jan.l 

5 

11 


Jan.l 

6 


James  '. 

Fowler,                  Cr. 

$    c.                                                          $    c. 

Iron  3Cwt. 

15 

60 

Jan.2  4 

Wheat  12  bushels, 

18 

50 

Kiim  10  galls. 

10 

20 

6 

5 

Corn       7     do. 

3 

71 

Wine  3  galls. 

0 

36 

10 

7 

Cash, 

7 

84 

Lot  Ford,  Dr. 

Contra     Cr. 

Broadcloth  2  yds. 

9 

50 

Jan.3 

4 

Beef  UOlbs. 

4 

40 

Wine  1  gall. 

2 

12 

7 

6 

Wood  3  cords, 

6 

00 

In  the  preceding  example,  the  two  columns  on  the  left  both  of 
the  Dr.  and  Cr.  accounts  contain  the  date  of  the  transaction  and 
the  page  of  the  Day  Book  on  which  the  original  account  is  to  be 
found.  Next  follows  the  article  and  its  quantity,  which  should  be 
written  in  few  words,  and  then  its  amount  in  the  money  columns. 
Either  the  date  or  the  page  of  the  Day  Book  which  contains  the 
account,  is  anjply  sufficient  in  the  Leger,  and  the  latter  is  to  be 
preferred. 

The  Leger  exhibits  at  one  view  the  accounts  with  an  individual, 
as  it  contains  on  the  Dr.  siiie  whatever  he  has  received,  and  on 
the  Cr.  side  whatever  he  has  paid.  The  difference  between  the 
sums  of  Dr.  and  Cr.  called  the  Balance,  shows  the  state  of  the  trade 
in  this  instance. 

An  Index  must  accompany  the  Leger,  in  which  the  names  are 
arranged  alphahetically,  with  the  page  of  the  Leger  on  which 
each  account  is  to  be  found.     See  the  Index  to  the  Leger  for  Ex.  2, 

The  following  general  directions  are  to  be  observed  in  forming 
the  Leger.  Let  each  account  be  posted  from  the  Day  Book  in  its 
proper  place  in  the  Leger.  If  a  mistahe  be  maile,  let  it  be  cor- 
rected by  an  account  in  the  Day  Book,  clearly  stating  the  correc- 
tion, and  then  let  this  account  be  posted  in  its  proper  place  in  the 
Leger,  that  no  blot  or  erasure  may  disfigure  its  pages. 

THE  CASH  BOOK. 

la  this  book  are  recorded  the  daily  receipt  and  payment  of  men 
ey.  For  this  purpose  there  are  two  polumns,  one  for  money  re- 
ceived, and  the  other  for  money  paid,  in  which  should  be  recorded 
merely  the  date,  to  or  by  whom  paid,  and  the  sum.  The  Cash 
Book  is  convenient,  but  not  absolutely  necessary.  By  some  ac- 
countants other  auxiliary  books  are  used,  which  are  found  to  be 
useful  or  imj)ortant  in  some  particular  business.  These  the  ac- 
countant will  readily  form  for  himself,  as  circumstances  may  render 
nef:es-ary. 

Note  1.  As  several  of  the  preceding  books  may  be  necessary 
in  the  progress  of  business,  they  should  be  distinguished  by  letter- 
ing them  in  the  following  manner.  Day  Book  A,  Day  Book  B,  &c. 
Leger  A.  Leger  B.  &c.  And  in  posting  accounts  into  the  Leger, 
there  must  be  a  reference  to  Day  Book  A.  or  B.  kc.  as  the  ac- 
count is  Ibuod  in  the  one  or  the  other.     See  Example  2. 


BY  SINGLE  ENTII.Y 


4in 


Note  2.  In  the  following  example  the  barter  of  any  article,  as 
well  as  the  sale  of  an  article  for  cash,  is  entered  into  the  Day 
Book,  although  such  accounts  are  not  to  be  posted  into  the  Leger. 
This  is  not  generally  practiced,  but  the  accountant  will  often  find  a 
material  benefit  in  recording  even  these  changes  of  property. 

EXAMI'LE  L     SINGLE  ENTRY. 
DAY  BOOK. 


January  1,  1820. 
My  whole  property  is  a  debt  of  400  dollars  due 
me  from  Samuel    Richards,  the  balance  of  my  in- 
heritance. 

$ 

c. 

Jan.l,  1820. 

Samuel  Richards,                   Dr. 

To  balance  Irom  the  estate, 

400 

^ 

3rd. 
Samuel  Richards,                    Cr. 
By  broadcloth  105  yards  at  3  dolls  per  yard, 

315 
192 

4th. 

John  Higgins,                         Dr. 

To  55  yards  of  broadcloth,  at  $3  50c.  per  yard, 

50 

5th. 
Exchanged  40yds.  of  broadcloth,  for  24Cwt.  Iron 

at  g5. 

6th. 

John  Higgins,                        Cr. 

By  cash,  on  account, 

180 

7th. 
Sold  S.  M.20Cwt.  of  Iron  at  gSiperCwt.  forcash. 

105 

Note.  The  preceding  example  contains  so  few  accounts,  that 
the  formation  of  the  Cash  Book  is  unnecessary.  It  is  sufficient, 
however,  to  illustrate  the  principles  of  Single  Entry ;  while  it  is 
so  short  that  the  whole  may  be  easily  comprehended  by  the  pupil. 
The  Leger,  in  which  this  Day  Book  is  posted,  is  on  the  following 
page. 


492 


BOOK  KEEPING 


EXAMPLE  I.    SINGLE  ENTRY. 
LEGER. 


Day 

Book 

1 


Dr. 
Samuel  Richards 
Balance  fromestate, 


John  Higgins,  Dr, 
Broadcloth,  55yds. 


i 

400 

192 

c. 

60 

Day 
Book 

1 
1 

Contra        Cr. 
Broadcloth  105yds. 
Balance, 


Contra 
Cash, 
Balance, 


Cr. 


$ 

c. 

315 

85 

400 

180 

12 

50 

— 

192 

50 

To  balance  the  accounts,  place  the  difference  of  the  several  ac- 
counts under  the  sQ^alier  side.  Thus  in  the  account  with  Samuel 
Richards,  §85  is  the  Cr.  to  balance  ;  and,  in  the  account  with  John 
Higgins,  $\^  50c.  is  the  balance  on  the  same  side.  It  is  obvious 
that  I  have  gained  by  the  trade.  Were  not  the  gain  evident,  on 
inspection,  it  would  be  made  so  by  the  following  inventory  from 
the  preceding  Leger. 


January  8,  1820. 
Due  from  Samuel  Richards, 

John  Higgins,  .         -         . 

On  hand  10yds.  broadcloth  at  jj3  per  yard, 

Cash  from  broadcloth  and  iron, 

4Cwt.  of  iron  at  g5     - 


My  property  Jan.  1, 


Amount, 


i  c. 

85  00 
12  50 
30  00 
285  00 
20  00 

432  50 
400  00 


Gain, 


32  50 


BY  SINGLE  ENTRY. 


19: 


EXAMPLE  II.    SINGLE  ENTRY 
DAY  BOOK  A. 


0 

January  1,  1821. 
Inventory -of  all  my  property  and  debts,  taken 
this  day  by  me,  A.  B. 

£      s.     d. 
Ready  money,                                      300     0     0 
House  and  Furniture,                          500     0     0 
Williams  Farm,                                   600     0     0 
Merchandise,                                        555     0     0 
Produce,                                                 45     0     0 

£ 

S' 

d. 

I  owe,  on  accounts, ' 
To  Henry  Hardy,          £15  10s. 
To  Thomas  Howe,          30     0 

2000     0     0 
45   10     0  . 

My  net  estate. 

1954   10     0 

January  1,  1821. 

Henry  Hardy, 
By  former  account,  balance 

Cr. 

15 

30 

9 
9 

18 

1 

2 
3 

1 
14 

16 

0 

18 

15 

15 
6 

"2 

5 

0 

16 

6 

0 

Thomas  Howe, 

By  former  account,  balance 

Cr. 

0 

Salmon  Rogers, 
To  20  bushels  of  wheat,  at  9s.  pe 
To  6  yards  of  broadcloth,  at  33s. 

Dr. 

;r  bushel, 
per  yd. 

0 

0 

~0 

John  Wheat, 
To  20  gallons  of  Rum,  at  6s.  9d. 

Dr. 

per  gall. 

0 

John  Taylor, 
To  61bs.  loaf  sugar,  at  2s.  7d.  pe; 
Igall.  of  Rum, 

Dr. 

^  ife 

6 
9 

"3 

2d. 

Simon  Pond, 

To  6  bushels  of  wheat,  at  9s.  per 

Dr. 

bushel. 

0 

John  Wheat, 

By  cash,  on  account  60s. 

Cr. 

0 

3d. 

Henry  Hardy, 
To  3  gallons  of  wine,  at  12s.  per 
30  bushels  of  wheat,  at  9s.  6d 

Dr. 

gall, 
per  bush. 

0 
0 

~0 

2) 


BOOK  KEEPING 


)                               January  3d,  1821. 

Titus  Coale,                           Dr. 

To  20  gallons  of  Rum,  at  7s.  per  gall. 

^cwt.  of  Havanna  sugar,  at  60s.  per  cwt. 

7 

0 
15 

0 
0 

7 
1 

15 
13 

0 

4th. 
John  Wheat,                           Cr. 

By  SOlbs.  of  nails,  at  8d.  per  ife 

4 

Simon  Pond,                           Cr. 

By  cash,  on  account  25s. 

1 

3 
3 

6 

16 
14 

0 

6th. 

Peter  Owen,                           Dr. 
To  goods  dehvered  C  Paige,  by  your  order, 

8 

Dixon  Ferry,                          Dr. 
To  66  yards  cotton  cloth,  at  Is.  4d.  per  yd. 

8 

8th. 
Henry  Hardy,                         Dr. 

To  2  yards  blue  broadcloth,  at  36s.  per  yd. 

3 

12 

0 

Salmon  Rogers,                     Cr. 

By  cash  to  balance, 

18 
20 

18 
6 

0 

Peter  Pindar,                          Dr. 

To  30  galls,  wine,  at  13s.  6d.  per  gall. 

0 

9th. 

Titus  Coale,                           Cr. 

By  25  bushels  wheat,  at  8s,  6d.  per  bush. 

10 

12' 

6 

Hervey  Brown,                       Dr. 
To  6yds.  hrowfi  linen,  at  Is.  9d.  per  yd. 

1 

15 
6 

9 

8 
14 

~3 

0 

16 

4 

9 

lOvh. 
John  Merrill,                           Dr. 
To  lOIfe  nails,  at  lOd.  per  ft 
16Jfe  brown  sugar,  at  lid. 

4 

8 

~0 

Thomas  Howe,                       Dr. 

To  cash,  on  account. 

4 

Titus  Coale,                           Dr. 

To  16galls.  Hum,  at  7s.  3d. 

0 

12th. 
Peter  Owen,                            Cr. 
By  20J-  bushels  wheat,  at  9s. 

6 

BY  SINGLE  ENTRY. 


49i 


Jannaiy  13th,  1821. 

Simon  Pond,                          Dr. 

To  3  gallons  Rum,  at  7s. 
l-^gall.  wine,  at  12s.  4d. 

1 

1 

18 

(3 

0 
6 

1 

19 

6 

14th. 

John  Wheat,                          Cr. 

By  cash  in  full,  4U.  8d. 

2 

J 

8 

17th. 

Thomas  Howe,                      Dr. 
To  20gaHs.  rum,  at  7s.  3d.  per  gall. 
5  do.  wine,  at  12s.  4d. 
40lbs.  nails,  at  9d. 

7 

3 

1 

11 

6 

1 

10 

16 

0 
8 

8 

18th. 

Charles  Gray,                        Dr. 
To  3  gaUons  French  Brandy,  at  12s.  per  gall. 
lOf  lbs.  loaf  sugar,  at  2s.  4d. 

1 
1 

16 
5 

0 

I 

3 

1 

1 

Dixon  Ferry,                          Cr. 
By  1  mahogany  table,  72s. 
1  wash  stand,  17s. 

3 

12 
17 

0 
0 

4 

9 

0 

19th. 
Simon  Pond,                           Cr. 

By  cash,  on  account  43s.  6d. 

2 

3 

6 

20th. 

Peter  Pindar,                           Cr. 

By  40  bushels  of  wheat  at  95. 
cash,  on  account  45s. 

18 

0 
6 

0 
0 

20 

1 

6 
2 

0 

23d. 

John  Taylor,                          Cr. 

By  20|Ife  butter,  at  Is-  Id. 

3 

Henry  Hardy,                         Dr. 

To  30  bushels  of  wheat,  at  9s.  66. 

14 

2 

2 

5 

8 

15 

3 

Is 

0 

26th. 
Hervey  Brown,                        Cr. 

By  cash,  on  account  8s.  9d. 

9 

Titus  Coale,                             Cr. 
By  6i  bushels  wheat,  at  8s.  6d. 
cash,  on  account  3s.  3d. 

3 

~6 

4} 


BOOK  KEEPING 


January '29th,  1821. 

Peter  Owen, 
To  Icvvt.  white  Havanna  sugar,  j 
lOgalls.  Rum,  at  8s. 

By  cash,  on  account  35s.  2(1, 

at  63s. 
Cr. 

Dr. 

3 
4 

7 

3 

0 

~3 

0 
0 

"o 

1 

15 

2 

Dixon  Ferry, 

To  20lbs.  brown  sugar,  at  1  hi. 

Dr. 

18 

4 

John  Merrill, 

By  cash  in  full  23s. 

Cr. 

1 

3 
3 

3 

3 
12 

0 

31st. 

Thomas  Howe, 

To  Icwt.  Havanna  sugar,  at  G3s. 

Dr. 

0 

Charles  Gray, 

By  ^  bushels  wheat,  at  8s.  Gd. 

Cr. 

3 

Samuel  Lyman, 
To  100  bushels  of  wheat,  at  9s. 

By  cash,  on  account  £40  10s. 

Cr. 

Dr. 

45 
40 

3 

0 

3 

0 

0 

Joshua  Noble, 

To  Icwt.  Havanna  sugar,  at  63s. 

Dr. 

0 

SINGLE  ENTRY 


497 


LEGER  INDEX, 

TORMZB  FOR   THE   FOLLOWING   LEGER,  INTO   WHICH   THE   PRECEPING  DAY 
BOOK   IS   POSTED. 

A.  B.  C. 

Brown,  Hervey  p.  2.         Coale,  Titus  p.  2. 


E.  F. 

Ferry,  Dixon  2. 


G.  H. 

Gray,  Charles  3.  Hardy,  Henry  1. 

Howe,  Thomas  1. 


K. 


L.  M. 

Lyman,  Samuel  3.  Merrill,  John  2. 


Noble,  Joshua  3. 

Owen 

,  Peter  2. 

Pindar,  Peter  2. 
Pond,  Simon  2. 

% 

Rogers, 

R. 

Salmon  1. 

3. 

T. 
Taylor,  John  1, 

u. 

v> 

W. 

Wheat,  John  1. 

X. 

y. 

Note.  In  the  following  Leger,  both  the  date  of  the  transaction 
and  its  page  on  the  Day  Book  are  given,  merely  to  exercise  the 
learner,  as  only  one  of  them  is  essential. 


P:; 


i9(} 


0) 


BOOK  KEEPING 
Example  H.     LEGER  A. 


1821. 

P. 

D.B. 

A. 

Jan.  3 

1 

8 

Q 

23 

3 

Jan.  10 

2 

17 

3 

31 

4 

Jan.  1 

1 

Jan.  1 

1 

Jan.  1 

1 

Henrj  Hardy, 

Wine  3  gallons  at  12s.* 
Wheat  30  bushels  at  9s.  6(1. 
Blue  broadcloth  2yds.  at  36s. 
Wheat  30  bushels  at  9s*  6d. 


Thomas  Howe, 

Cash, 

Rum  20  gallons  at  7s.  3d. 
Wine   6    do.  ISs.  4d. 

Nails  40lbs.  9d. 

Havana  Sugar  iCwt. 


Salmon  Rogers, 

Wheat  20  bushels, 
Broadcloth  6yds. 


John  Wheat, 


Rum  20  gallons, 


John  Tajlor, 

Loaf  Sugar  6ibs. 
liurn  1  gallon, 


Dr. 


Dr. 


Dr. 


Dr. 


Dr 


£ 

s. 

1 

16 

14 

5 

3 

12 

14 

6 

33 

18 

15 

0 

7 

5 

3 

1 

1 

10 

3 

3 

30 

0 

16 


16 


*  In  posting  accounts  into  the  Leger,  some  accountants  write  tlie  article,  quan- 
l.i(y,Hnd  price,  as  is  here  done  ;  others  omit  the  price  in  the  Leger,  as  the  col- 


BY  SINGLE  ENTRY. 
Example  II.     LEGER  A. 


499 
(1) 


1821. 
Jan,  1 


Jan.  1 


Jan.  8 


Jan.  2 

4 

14 


Jan.  23 


P. 

D.B 

A. 

I 


Contra 

Balance  of  former  account, 
Balance, 


Contra 


Cash, 


Contra 

Cash, 

Nails  50lbs. 
Oash, 


,  Contra 

3        Butter  20iib, 


Cr. 


Contra  Cr. 

Balance  of  former  account. 


Ur. 


Cr. 


Cr. 


10 


^)^  li     0 


30 


18 


3 

0 

1 

13 

2 

1 

6 

15 

rectness  of  each  account  is  to  be  ascertained  in  theDay  Book.    The  latter  meth- 
od is  sufficient,  and  is  generally  followedl  m,  this  Exampl«. 


5d(; 


(2) 


BOOK  KEEPING 
Example  H.     LEGER  A. 


1821. 

Jan.  2 

13 

P. 

D.B. 

A. 

1 
3 

2 

2 

2 
4 

2 
4 

2 

2 
2 

Simon  Pond,               Dr. 

Vheat  5  bushels, 

Rum  3  galls.=21s.  and  Winel|  gall.= 
IBs.  6d. 

2 

1 
4 

7 
5 

1 

s. 
6 

19 
-4 

16 

16 

6 
6 

Jan.  3 
10 

Titus  Coale,                Dr. 

Rum  20galls.=140s.  and  Sugar  |Cvvt.= 

16s. 
Rum  16  galls. 

0 
0 

13 

11     0 

Jan.  6 

29 

Peter  Owen,               Dr. 

Goods  per  your  order  to  C.  Paige, 
Havanna  Sugar,  lcwt.=63s.  and  Rum  10 
galls. =80s. 

3 

7 
10 

16 

3 

8 

0 
~8 

Jan.  6 

29 

Dixon  Ferrj,             Dr. 

Cotton  Cloth  66yds. 
Brown  Sugar  201bs. 

3 

„  ,  ,. 
4 

20 

14 

18 

Ts 

5 

8 
4 

0 

Jan.  8 

Peter  Pindar,              Dr. 

Wine  30  gallons, 

0 

Jan.  9 

Hervey  Bi^own,          Dr. 

Brown  Linen,  ^yds. 

8 

9 

Jan. 10 

John  Merrill,              Dr. 

Nails  lOlb. 
Brown  Sugar  161b. 

8 
14 

4 

8 

1    1    o\ 

0 

,nkt   , 

BY  SINGLE  ENTRY. 
Example  II.     LEGER  A. 


sei 


<2) 


1821. 

Jan.  4 

19 

P. 

D.  B. 

A. 

2 
3 

Contra                 Cr. 

Cash, 
Cash, 
Balance^ 

1 
2 

■    4 

10 

2 

13 

s. 

5 

3 

16 

"4 

12 

15 

3 

IT 

0 
6 

^> 
0 

Jan.  9 

26 

2 
3 

2 
4 

3 
3 

Contra                Cr. 

Wheat  25  bushels, 
Wheat  61     do. 
Cash, 

6 
3 
3 

~0 

Jan.  12 

29 

Contra                 Cr. 

Wheat  20i  bushe!s, 
Cash, 

9 

1 

4 

15 

6 

2 

10 

IS 

0 

Jan.  18 

Contra                 Cr. 

1    Mahogany   Table  72s.  and    1    Wash 

Stand  17s. 
Balance, 

4 
4 

9 

4 

13 

"0 

0 

Jan.  20 

Contra                 Cr. 

Wheat  40  bushels=£i8  and  cash  £2  5s. 

20 

5 

0 

Jan,  26 

3 
4 

Contra                 Cr. 

Cash, 

1 

0 
0 

3 

9 

Jan.  29 

'      Contra                 Cr. 

Cash, 

• 

0 

502 
(3) 


BOOK  KEEPING 
Example  II.     LEGER  A. 


1821. 
Jan.  18 

P. 

D.  B. 

A. 

3 

Jan.  31 

4 
4 

Jan.  31 

Charles  Gray, 

French  Brandy,  3  gallons, 
Loaf  Sugar,  lO^lbs. 
Balance, 


Joshua  Noble, 

Havanna  Sugar,  Icwt. 


Dr. 


Samuel  Ljman,        Dr. 

Wheat  100  bushels, 


Dr. 


d. 


45 


3    0 


BY  SINGLE  ENTRY. 
Example  II.     LEGER  A. 


503 
(3) 


1821. 
Jan.  31 

P. 

D.  B. 

A. 

4 

Contra                Cr. 

Wheat  8J  bushels, 

3 

40 
4 

45 
3 

s. 
12 

10 
10 

"o 

3 

d. 
3 

Jan. 31 

4 

Contra                Cr. 

Cash, 

Balance, 

0 
0 

~0 

Contra                Cr. 

Balance, 

0 

CASH  BOOK, 


FORMED   FROM   THE   DAY    BOOK,  EXAMPLE   II. 


'Jan.  2. 

4. 

8. 

14. 

19. 

.  20. 

^26. 

29. 

31. 


Received. 

Of  John  Wheat,  3 


S.  Pond, 
S.  Rogers, 
J.  Wheat, 
S.   Pond, 
P.  Pindar, 
H.  Brown, 
T.  Coale, 
P.  Owen, 
J.  Merrill, 
S.  Lyman, 


1 

18 

2 
o 


s. 
0 
5 

18 
1 
3 
5 
8 
3 

15 
3 


40   10 

0 

73  13 
24     4 

4 
8 

49     8 

8 

300     0 

~0 

349     8 

8 

^      Paid. 

£   s.  d. 
Jan.  10,  Thomas  Howe,    15  0  4 

To  this  add  my  ex- 
pences  for  the  month, 
being  9  4  4 

Amount  paidl^  24  4  8 


Amount  received, 
do.      paid. 

Excess, 

Cash,  Jan.  1. 

Do.  Feb.  1. 


In  balancing  the  Leger,  as  in  Example  2,  draw  two  licavy  black 
lines  under  the  accounts  in  which  the  sums  of  Dr.  and  Cr.  are 
equal,  to  show  that  the  account  is  settled,  'Where  the  sums  of  Dr. 
and  Cr.  are  unequal,  place  the  sum  to  balance,  under  the  smaller 


m4 


BOOK  KEEPING 


account,  nriting  against  it  the  word,  Balance.  As  there  is  to  be 
no  line  drawn  under  these  accounts,  and  as  there  is  no  reference 
HI  the  marginal  columns  to  the  Day  Book,  it  will  be  obvious  on  in- 
spection, that  such  accounts  are  not  settled. 


TO  FIND  THE  PROFIT  OR  LOSS, 

An  Inventory  of  the  property  and  of  the  debts  must  be  taken,  as 
follows,  from  Example  2. 

February  1,  1821. 

Inventory  of  all  my  property,  and  of  the  sums  doe  to  me,  or 
owed  by  me,  taken  this  day  by  me  A.  B. 

£        3.     d. 
Ready  Money,  349     8     8 

House  and  Furniture,  604     9     0 

Williams  Farm,  600     0     0 

Merchandise,  480     0     0 

Produce,  10     0     0 


1943  17 

8 

£ 

s. 

d. 

Due  from  Henry  Hardv, 

18 

8 

0 

S.  Pond, 

16 

0 

D.  Ferry,      H 

4 

0 

S.  Lyman, 

4 

10 

0 

J.  Noble, 

3 

3 

0 

27 

1 

0 

i  owe  Charles  Gray, 

11 

3 

26     9 

Difference, 

26 
Feb.  1. 

9 

9 

9 

Net  amount  of  property, 

1970     6 

8 

do. 

Jan.  1. 

1954  10 

0 

Profit  in  the  month, 


15  IS     8 


GENERAL  REMARK  ON  SINGLE  ENTRY. 

Book  Keeping  by  Single  Entry,  shows  clearly  the  state  of  ac- 
counts with  individuals,  but  it  does  not  exhibit  the  true  state  of  his 
affairs  to  the  book  keeper  himself.  For  this  purpose,  he  must  take 
an  Inventory  of- all  his  property  and  debts,  to  ascertain  the  quan- 
tity of  goods  unsold,  and  the  net  amount  of  his  property,  and 
thence  the  profit  or  loss  of  trade,  in  the  manner  just  taught.  This 
is  a  work  of  much  difficulty  and  trouble,  if  the  business  be  exten- 
sive. It  is  for  this  reason,  thut  book  keepit>g  by  Double  Entry  is 
preferred  in  extensive  trade. 


BY  SINGLE  ENTRY« 


505 


SHORTER  METHOD  OF  POSTING  ACCOUNTS  IN  SINGLE  ENTRY. 

The  method  a-lready  giren,  of  posting  accounts  from  the  Day 
Book  into  the  I.eger,  is  g«?neraily  considered  the  most  correct. 
The  following  shorter  method  is  perhaps  more  commonly  used. 

The  Leger  is  ruled  as  before,  and  merely  the  amount  of  an  ac- 
count is  posted  into  the  Leger,  preceded  by  the  page  and  letter  of 
th6  Day  Book  on  which  the  account  is  found.  The  first  two  ac- 
counts of  the  preceding  Leger  are  here  posted  from  the  Day  Book 
for  ah  example  of  this  method. 

LEGER.  SHORTER  METHOD. 


Dr.     Heiirj 

A  2.  £16  Is.  and £3  12 
A  4.  £14  5s. 

£ 

19 
14 

33 

15 
11 
_3 

30 

s. 

13 

5 

To 

0 

16 

3 

d. 

0 
0 

4 
81 
0 

~0 

Hardj, 

A  1.£15  10s. 
Balance, 

Cr. 

£ 

15 
18 

33 

s. 

10 

8 

T8 

d. 

0 
0 

Dr.    Thomas 

A  3.  £1^5  Os.  4d.  and 

£11    16s.  8d. 
A  4.  £3  3s. 

Howe, 

lA  l.£30. 

Cr. 

30 

0 

0 

In  this  Leger,  A  1,  A  2,  &c.  means  that  on  page  1,  2,  &c.  of  Day 
Book  marked  A,  that  particular  account  is  to  be  found. 

The  manner  of  balancing  the  Books,  and  of  ascertaining  the 
Profit  or  Lo?s  is  the  same  as  before  taught.  To  make  the  subject 
familiar  the  learner  should  be  directed  to  form  a  Day  Book  for 
himself,  and  to  carry  the  accounts  through  the  several  books,  ac- 
cording to  the  preceding  principles. 

SHORTEST  METHOt)  OF  KEEPING  ACCOUNTS. 

Only  one  Account  Book  is  necessary  in  the  practice  of  this  meth- 
od. It  is  formed  precisely  like  the  Leger  in  Single  Entry,  except 
that  there  is  no  column  of  reference  to  any  other  book.  The 
transactions  of  trade  are  entered  under  the  names  against  the 
date  on  v\rhich  they  take  place.  An  alphabet  for  the  arrangement 
of  the  names  is  found  convenient  for  reference  to  the  various  ac- 
counts. 

This  Account  Book  is  designed  to  answer  the  double  purpose  of 
Day  Book  and  Leger.  if  the  person  be  careful  to  enter  every 
instance  of  debt  and  credit  at  the  time  it  occurs,  he  will  be  able  to 
ascertain  at  any  time  the  state  of  his  accounts  in  a  particular  case. 
This  is  the  great  object  of  this  metho?!;  'vhich  is  exhibited  io  the 
feliowinpr  example, 

Q3 


6Qi^ 


BOOK  KEEPING 


ACCOUNT  BOOK. 


i 


Linden,  Jan.  1,  1820. 

— =-i 

182G. 

John  Wilson j                         Dr. 

$ 

c. 

Jan.  1 

To  3  cords  of  wood,  ^1   50  per  cord, 

4 

60 

6 

To  li  days  work,  by  hired  man,  at  66  cents, 

1 

00 

9 

To  6i  bushels  rye,  at  60  cents, 

2 

75 

13 

To  3  bushels  wheat,  at  $1   76, 

5 

26 

Feb.  1 

To  5  cords  wood,  at  gf  60, 

^  7 

50 

March  9 

To  7  bushels  oats,  at  31  cents. 

2 

17 

16 

To  work  with  hired  man  and  horses,  one  day, 

1 

66 

24 

83 

18  Cash  to  balance, 

34 

25 

17 

1820. 

Peter  Pajwell,                     Dr. 

July  1 

To  life  hyson  tea, 

1 

66 

9 

To  lOlbs.  brown  sugar,  at  19  cents, 

1 

90 

26 

To  3  gallons  Rum,  at  $1   17, 

3 

61 

Sept,  9 

To  91bs.  blister  steel,  at  10  cents, 

90 

11 

To  6|yds.  calico,  at  54  cents. 

3 

51 

Oct  3 

To  3vds.  cotton  cloth,  at  18  cents, 

0 

54 

13 

To  2lbs.  loaf  sugar,  at  31  cents, 

0 

62 

1821. 

Jan.  1 

To  life  hyson  tea, 

1 

46 

3 

To  goods  delivered  by  your  order  to  E,  T. 

3 

72 

V7 

72 

BY  SINGLE  ENTRY. 

SHORTEST  METHOD. 


-607 


1820. 

Jan.  1 

5 

13 

Feb.  1 

March  7 


Linden,  Jan.  1,  1820. 

John  Wilson,  Cr. 

By  1216  shingle  nails,  at  10  cents, 

By  25ilbs.  cheese,  at  7  cents, 

By  your  order  on  John  Gibbs,  for  goods, 

By  20ilbs.  Butter,  at  18  cents, 

By  licwt.  iron,  at  ^6  60  per  cwt. 


77P 


c. 
20 
78 

75 
69 
75 
17 


1820. 

Contra.                           Cr. 

Aug.  4 

By  121bs.  butter,  at  121  cents, 

1 

50 

7 

By  work  2  days  by  your  man. 

1 

33 

Oct.  3 

By  661bs.  chees€,  at  7  cents. 

3 

92 

10 

By  cash, 

3 

00 

1821. 

Jan.  1 

By  12  bushels  rye,  at  50  cents, 

6 

00 

3 

By  cash  to  balance, 

1 

97 

17 

72 

m^ 


BOOK  KKEPINQ 


BOOK  KEEPING  BY  DOUBLE  ENTRY  * 

THIS  npethod  of  Book  Keeping  differs  from  that  by  Single  Entry 
in  t\^o  important  respects,  viz,  things,  as  well  as  persons,  are  en- 
tered as  Dr.  and  Cv.  and  as  Dr.  and  Cr.  to  each  other,  and  each 
account  is  entered  twice  in  the  Leger.  For  the  latter  particular 
it  is  called  Double  Entry. 

In  the  practice  of  Double  Entry,  three  Principal  Books,  and  four 
Auxiliary  Books  are  neceseary. 

PRINCIPAL  BOOKS. 
The^eare  the  Day  Book  or  Waste  Book,  the  Journal,  and  the  Leger. 

^  L     THE  DAY  BOOK  OR  WASTE  BOOK. 

IpjfP'  The  Day  Book  beginj^,  as  in  Single  Entry,  with  an  Inventory  oX 
^  '"  all  the  property  and  (iebts  of  the  merchant,  and  is  followed  by  a 
regular  account  of  Ib^  transactions  in  business,  in  the  order  of 
time  in  which  they  occur,  stated  in  language  so  explicit  and  ful8 
that  there  can  be  no  mii<take.  'i'his  book  is  in  Double  JLntry,  a 
ip^re  record  of  the  changes  of  i*'rapprty,  ^nd  Dr.  and  Cr.  are  riot 
introdaced  into  it-  References  are  made  in  it  to  the  auxiliary 
boioki^,  when  it  is  necessary.     It  is  ruled  as  in  Single  entry. 

To  exhibit  the  diilerence  in  the  t^yo  methods  of  Book  Keepings 
the  principles  of  Double  Entry  will  be  iU,ustra>ted  by  Exs^mple  I. 
of  Single  Entry. 

(1)  EXAMPLE  I.  DOUBLE  ENTRY.  DAY  OR  WASTE  BQOK. 


January  1,  1821. 
My  whole  property  is  a  debt  of  ^400  due  me  from 

Samuel  Richards,  the  balance  of  my  inheritance. 
Samuel  Richards  owes  me  the  balance  of  my  inher- 

ifance, 

400 

c. 

ord. 
Bought  of  Samuel  Richards  105  yards  of  broadcloth 
at  3  dolls,  per  yard. 

315 

4th. 
Sold  John  Higgins  55yds.  of  broadcloth  at  $3  50c. 
per  yard. 

192 

50 

5lh. 
Bartered  40yds  of  broadcloth  for  24cwt.  of  Iron,  at 
5  dolls.  |)<^r  Cwt. 

120 

6th, 
Received  of  John  Hi^jjins  in  part. 

180 

7th. 
Sold  20cwt.    of  Iron   to  S.  M.  for  Cash  at  51  dolls, 
per  Cwt. 

105 

*  The  general  principles  of  this  system  of  Book  Keeping  are  taken  from  the  Sys- 
tem in  Rees' Cyclopedia,  which  is  generally  adopted  by  the  merchants  of  London. 


BY  DOUBLE  ENTRY. 


500 


Note.  In  recording  transactions  in  the  Day  Book,  the  above 
order  is  generally  to  be  preserved,  viz.  1st  the  date  ;  2d  the  kimi 
of  transaction  in  the  active  voice,  as  owes,  s61d,  bought,  exchanged, 
&c. ;  3d  the  name  of  the  person  ;  4th  the  article  and  quantity  ;  6th 
the  price  ;  and  6th  the  amount  in  the  columns  of  money. 

II.    THE  JOURNAL. 

The  object  of  the  Journal  is  to  prepare  the  accounts  for  thf 
Leger.  To  effect  this,  the  Dr.  and  Cr.  of  every  article  contained 
in  the  Day  or  Waste  Book,  is  ascertained  and  expressed  in  the 
Journal. 

The  Journal  is  ruled  v^^ith  two  blank  columns  on  the  left,  viz, 
one  for  the  date  and  the  other  for  the  page  in  the  Leger,  and  with 
the  proper  columns  for  money  on  the  right,  as  in  the  following 
journal  of  the  preceding  Day  Book. 

EXAMPLE  I.  DOUBLE  ENTRY. 


JOURNAL. 

0^ 

^     - 
.  Jan. 
1 

p 
L. 

yamuel  Kictiards  Dr.  to  :5tock       §400  OU 
For  the  balance  of  my  inheritance, 

400 

0 

3 

Broadcloth  Dr.  to  Samuel  Richards  315  00 
For  105yds.  broadcloth  at  3  dolls,  per  yd. 

315 

4 

John  H!£,-irins   Dr.  to   broadcloth       192  50 
For  55ydsr broadcloth  at  $3  50  per  yd. 

192 

50 

6 

Iron  Dr.  to  broadcloth                       120  00 
To  40yds  broadcloth  at  3  dolls,  per  yard  ) 
for  24cwt.  of  iron,                                       ^ 

120 

^- 

6 

'Cash  Dr.  to  John  Higgins                  180  00 
Received  of  him  oh  account, 

180 

7 

Cash  Dr.  to  Iron                           ^        105  00 
Received   for  20cwt.  of  iron  at  J5  25  per 
Cwt. 

105 

To  understand  the  method  of  forming  the  Journal,  the  following 
distinctions  must  be  attendecVto.  Accounts  are  distinguished  in- 
to personal,  real,  and  fictitious.  Personal  accounts  are  those  in 
which  a  person  is  entered  as  Dr.  or  Cr.  and  are  the  same  as  in 
Single  Entry.  Thus,  in  the  preceding  Journal,. Samuel  Jiichards 
is  Dr   in  one  account. 

Real  accounts  are  those  of  property  of  any  kind,  as  cash,  hous- 
es, cloth,  furniture,  adventure,  5fc.  In  the  preceding  example 
Broadcloth  is  Dr.  to  Samuel  Richards,  andiron  Dr.  to  broadcloth, 
kc. 

Fictitious  accounts  are  those  of  stock,  and  profU  and  loss.  Stoclg 
is  used  for  the  ovvner  of  tjie  bouks.     In  the  preceding^  Example, 


5ia 


BOOK  KEEPING 


Samuel  Richards  is  Dr.  to  stock,  i.  e.  to  the  owner  of  the  books. 
Profit  and  loss  is  used  for  either  gain  or  loss  in  the  course  of  trade. 
This  account  does  not  appear  in  the  Journal  but  in  the  Leger. 

If  two  or  more  persons  or  things  are  included  in  the  same  account 
in  the  Journal,  they  are  expressed  by  the  term,  Sundries. 

Rules  for  distinguishing  Dr.  and  Cr.  in  the  Journal,  are  the  fol- 
lowing. 

The  person  to  whom  or  for  whom  I  pay,  or  whom  I  enable  to 
pay,  is  Debtor. 

The  person  for  whom  or  from  whom  I  receive,  or  by  whom  I 
am  enabled  to  pay,  is  Creditor. 

Whatever  comes  into  my  possession  or  under  my  direction,  is  Dr. 

Whatever  passes  out  of  my  possession  or  from  under  my  contrqul, 
is  Cr. 

The  phrases,  In  debtor,  and  Out  creditor,  briefly  express  the 
points  in  these  rules.     Thus,  in  the  preceding  Journal,  Broadcloth, 


EXAMPLE  I.    DOUBLE  ENTRY. 
LEGER. 

Date 
1820. 

P. 
J. 

Stock,                             Dr. 

P. 

L. 

$ 

400 

315 

27 

342 

c. 

50 
50 

50 

Jan. 
1 

3 

1 
I 

Samuel  Richards,          Dr. 

To  Stock, 

Broadcloth,                    Dr. 

To  Samuel  Richards,  105yds.  at  ^3  per  yard, 
Profit  and  loss. 

4 
5 

1 
1 

John  Higgins,                 Dr. 

To  broadcloth,  SSyds.  at  $S  50, 

192 

Iron^                               Dr. 

To  broadcloth,  24cwt.  at  g3. 
Profit  and  loss, 

120 
5 

125 

7 
8 

1 

Cash,                              Dr. 

Po  John  Higgins, 
Iron,  20cwt.  at  $5\, 

180 
105 

285 

Profit  and  Loss,           Dr. 

m^ 


BY  DOUBLE  ENTRY. 


511 


Ithich  come«  into  my  possession,  is  Dr.  to  Samuel  Richards  by 
whom  it  is  paid  in  part  for  stock  in  bis  hands  ;  John  Higgins  is  Dr. 
to  broadcloth  ;  Iron  is  Dr.  to  broadcloth,  and  so  on. 

III.    THE  LEGER. 

The  object  of  the  Leger  is  the  same  as  in  Single  Entry.  But  as 
things,  as  well  as  persons,  are  introduced  in  the  Journal,  they  must 
have  separate  accounts  in  the  Leger  also,  where  the  respective 
Drs.  and  Crs.  are  to  be  arranged  under  their  respective  heads. 

The  Legerjis  ruled  with  columns  for  the  denominations  of  money 
on  the  right  side,  immediately  before  which  is  a  column  for  refer- 
ence to  the  page  of  the  Leger  in  which  the  corresponding  account 
is  found  ;  and  on  the  left  side,  is  a  column  for  dates,  and  another 
for  the  page  of  the  Journal  in  which  the  account  may  be  found. 
The  following  Leger  for  the  preceding  Example  is  formed  on  this 
plan.     See  foot  of  this  and  the  preceding  page. 


EXAMPLE  I.     DOUBLE  ENTRY. 
LEGER. 

Date.  P. 
1820.    J. 
Jan.  1      1 

Contra                     Cr. 

By  Samuel  Richards, 

P. 
L. 

$ 
400 

c. 

3 

1 

Contra                     Cr. 

By  Broadcloth, 

1 
1 

315 

4 
6 

1 
1 

Contra                    Cr. 

By  John  Higgins,  55yds.  at  $3  50, 
Iron,  40yds.  at  $3 

192 
120 

50 

6 

b 

1 
1 

Contra                    Cr. 

By  cash, 

1 

180 
105 

— 

Contra                     Cr. 

By  cash,  for  20cwt.  at  $5  25, 

Contra                    Cr. 

Contra                     Cr. 

By  cloth, 
Iron, 

27 

6 

32 

50 
50 

12  BOOK  KEEP li^G 

In  polling  the  Journal  to  form  Iho  preceding  Leger,  Samut'! 
Richards  is  posted  Dr.  tu  Stock  J5400,  and  Stock  is  posted  Cr.  by 
Samuel  Richards  for  the  same  sum  ;  Broadcloth  is  then  entered  Dr. 
ill  the  next  account  to  Samuel  Richards,  and  Samuel  Richards  Cr. 
by  broadcloth  ;  John  Higgins  is  posted  Dr.  to  broadcloth,  and 
Broadcloth  Cr.  by  J  Higgins,  and  so  on.  It  is  obvious  that  each 
Dr.  must  have  a  Cr.  and  each  Cr.  a  Dr.  and  that  every  transaction 
relating  to  any  one  account,  whatever  may  be  its  place  in  one  ac- 
count in  the  Loger,  niust  be  posted  also  on  the  proper  side  under 
the  head  to  which  it  belongs.  Thus,  while  Cash  is  Dr.  to  J.  Hig- 
gins in  the  6th  account  for  ^180,  J.  Higgins  is  Cr.  by  cash  for  the 
same  sum  in  4th  account;  and  while  Cash  is  Dr.  also  to  iron,  Iron 
is  Cr.  by  cash  to  the  same  amount,  in  the  6th  account. 

On  inspecting  the  preceding  Leger  it  is  evident,  that  in  the  per- 
sonal accounts,  as  those  of  S.  Richards  and  J.  Higgins,  all  the  arr 
tides  for  which  they  are  indebted  are  posted  on  the  Dr.  side,  and 
all  the  articles  they  pay  are  on  the  Cr.  side  of  the  account ;  in  the 
real  accounts,  as  those  of  broadcloth,  and  iron^  the  quantity  bought 
is  posted  on  the  Dr.  side,  and  the  quantity  soid  on  the  Cr  side  so 
that  the  quantity  unsold  and  the  profit  or  loss  may  be  readily  as- 
certamed.  In  the  fictitious  account  of  Profit  and  Loss,  the  loss  is 
to  be  posted  on  the  Dr.  side,  and  the  profit  on  the  Cr.  side,  so  that 
the  difference  must  show  the  net  gain  or  loss,  by  which  the  stock 
has  been  increased  or  diminished  in  the  course  of  trade. 

Having  ascertained  that  the  accounts  have  been  correctly  posted, 
the  next  step  is  to  balance  the  Leger.  This  is  to  be  don#  in  th^ 
following  matiner.  To  show  this  method  more  clearljs  the  pre- 
ceding Leger  is  repeated,  and  the  several  steps  m  balancing  the 
accounts  are  subjoined. 

I'd  the  preceding  Leger  subjoin  a  new  account.  Balance  Dr.  and 
Cr.  Begin  with  the  next  account  to  Stock,  and  place  the  balance 
of  Dr.  and  Gr  under  the  smaller,  to  make  them  equal,  viz.  ^85 
on  the  Cr.  side.  Put  this  sum  on  the  Dr.  side  of  the  account,  Bal- 
ance. For  if  S.  Richards  is  Cr.  by  Balance  85  dolls,  then  Balance 
must  be  Dr.  to  S.  Richards  for  the  same  sum.  Next,  lo  balance 
the  Broadcloth  account,  the  quantity  on  the  two  sides  mu5t  first  be 
made  equa},  and  the  value  of  the  unsold  cloth,  at  first  cost, be  placed 
\inder  the  amount  sold,  viz.  10  yards  at  jJS,  amoufiting  to  ^oO.  Thp. 
whole  sum,  viz.  |J342  50  must  be  equal  to  the  amount  bought  afid 
the  profit  on  that  sold.  Then  «J30  must  be  placed  ou  the  Dr.  side 
of  Balance,  for  if  Broadcloth  be  Cr.  for  the  balance  unsold,  then 
Balance  must  be  Dr.  for  the  same  sum.  Proceed  in  this  manner, 
through  all  the  personal  and  real  account?. 


BY  DOUBLE  ENTRY.  513 

Proof.* 

Having  balanced  all  the  accounts  except  those  of  Stock,  Profit 
and  Loss,  and  Balance,  in  the  first  place,  close  the  account  of  Prof- 
it and  Loss,  by  making  it  Dr  to  the  stock  gained,  viz.  ^32  50,  and 
then  make  Stock  Cr.  by  the  same  sum.  Next,  let  Stock  be  bal- 
anced by  the  necessary  sura,  viz.  ^432  50,  and  Balance  be  made 
Cr.  by  the  same  sum.  If  the  sides  of  the  account,  Balance,  are 
now  equal,  the  work  is  right. 

This  proof  is  complete  from  this  consideration.  By  this  method, 
the  cash  in  hand,  the  debts  due,  and  the  goods  unsold,  are  contain- 
ed on  one  side,  and  what  is  owed,  is  contained  on  the  other  side. 

Another  method  of  proof  is  to  add  the  profit  to,  or  subtract  the 
loss  from,  the  original  stock,  and  the  sum  or  difference  placed  to 
the  Cr.  of  Balance,  will  be  equal  to  the  &um  on  the  Dr.  side  of 
Balance. 

General  Remark. 

The  Journal  should  be  kept  up  with  the  Day  Book,  and  the  ac- 
counts should  be  regularly  posted  into  the  Leger,  that  the  books 
may  be  as  nearly  even  as  possible.  And  at  the  end  of  every  month, 
the  balance  should  be  made,  the  Journal  and  Leger  having  been 
carefully  examined  to  see  that  all  the  records  of  the  Day  Book  have 
been  carefully  transferred  into  the  Journal,  and  correctly  posted 
from  the  Journal  into  the  Leger. 

The  following  Example  is  sufficient  to  exhibit  the  principles  of 
Double  Entry.  It  was  designed  to.be  so  short  that  the  student 
might  have  the  whole  before  him  at  one  view.  It  is  too  short, 
however,  to  render  any  of  the  auxiliary  books  necessary.  In  ex- 
tensive business,  however,  these  books  are  essential.  In  the  prac- 
tice too  of  Double  Entry,  the  work  will  be  shortened  by  forming 
the  Journal,  in  the  manner  shewn  in  the  next  Example.  An  ac- 
count of  the  auxiliary  books  will  afterward  be  given,  and  a  speci- 
men of  each  one,  as  connected  with  the  next  Example  of  Double 
Xntry. 

*  See  pages  514  and  615. 


R 


514 


BOOK  KEEPING 


EXAMPLE  I.    DOUBLE  ENTRY. 
LEGER. 

Date. 
1820. 

Jan. 
1 

J. 

Stock,                             Dr. 

By  Balance,  my  net  estate, 

P. 
L. 

1 

1 

1 



1 

$ 
432 

400 

315 

27 

342 
192 

120 
5 

c. 
50 

50 
50 

50 

Samuel  Richards,          Dr. 

To  Stock, 

3 

4 
5 

■ 

Broadcloth,                    Dr. 

To  Samuel  Kichards,  105yds.  at  ^3  per  yard, 
Profit  and  loss, 

John  Higgins,                 Dr. 

To  broadcloth,  55yds.  at  ^3  50, 

Iron,                              Dr. 

To  broadcloth,  24cwt.  at  ^5, 
Profit  and  loss, 

1 
1 

125 

180 
105 

285 

7 

i 

! 



Cash,                              Dr. 

To  John  Higgins, 
Iron,  20cwt.  at  g5i, 

Profit  and  Loss,           Dr. 

To  stock  gained. 

i 

32 

50 

Balance,                         Dr. 

By  Samuel  Richards, 
Broadcloth,  unsold, 
John  Higgins, 
Iron,  unsold. 
Cash, 

85 
30 
12 
20 

205 

43!:^ 

50 
50 

IBY  DOUBLE  ENTRY, 


515 


EXAMPLE  I.    DOUBLE  ENTRY. 

LEGER. 

t 

Date. 
1820. 
Jan.  1 

3 

P. 
J. 

1 

1 

Contra                     Cr. 

By  Samuel  Richards, 
Profit  and  loss, 

P. 
L. 

1 
1 

$ 

400 
32 

432 

315 
85 

400 

c. 

50 

50 

50 
50 

Contra                    Cr. 

By  Broadcloth,  105  yards,  at  g3, 
Balance, 

1 

4 

6 

6 

1 
1 

Contra                    Cr. 

By  John  Higgins,  55yds.  at  $3  50, 
Iron,                         40yds.  at  $3 
Balance,                  10yds.  unsold,  at  $3 

lU5yds. 

1 
1 

1 

1 
1 
1 

192 

120 

30 

342 

Contra                    Cr. 

By  cash. 
Balance, 

180 
12 

192 

50 
50 

7 
1 

1 

Contra                     Cr. 

By  cash,  for  20cvvf.  at  $5  25, 
Balance,          4cwt.  at  ^5, 
24owt. 

105 

20 
125 

285 

27 

5 
-32 

50 

50 

Contra                    Cr. 

By  balance, 

Contra                    Cr. 

By  cloth, 
Iron, 

Contra                    Cr. 

By  slock,  my  net  estate, 

1 

432 

50 
1 

516  BOOK  KEEPING 

AUXILIARY  BOOKS.  ^^^|^ 

These  are  1,  the  Cash  Book  ;  2,  the  Bill  Book  j  3,  the  Invoice 
5ook  ;  and  4,  the  Sales  Book. 

These  books  are  important  to  the  accountant,  as  a  record  of 
particular  transactions  referred  to  in  the  Day  Book,  and  as  origin- 
al and  particular  records  of  those  transactions.  They  aid  especial- 
ly in  posting  accounts  into  the  Leger.  They  may  be  considered 
as  a  kind  of  Day  Book,  in  aid  of  the  general  Day  Book,  and  it  is 
obvious,  that  if  all  the  particular  accounts  were  arranged  under 
general  heads  in  separate  baoks,  the  common  Day  or  Waste  Book 
would  be  unnecessary,  except  as  exhibiting  a  general  history  of 
the  changes  of  property. 

I.    THE  CASH  BOOK. 

The  Cash  Book  is  a  record  of  all  money  paid  or  received.  It 
i«  referred  to  in  the  Day  Book,  by  the  initials  C.  B.  It  is  formed 
like  the  Leger,  with  a  Dr.  and  Cr.  side,  the  Dr.  side  containing  all 
money  received,  and  the  Cr.  all  money  paid.  The  transactions 
are  to  be  regularly  entered  into  the  Cash  Book  as  they  take  place, 
with  the  dates,  names,  and  all  necessary  particulars. 

The  man  of  business  will  find  it  convenient  to  have  separate 
columns  for  some  transactions,  as  of  money  accounts  at  a  Bank, 
Brokers,  kc.  and  for  some  small  incidental  expenses,  as  well  as  for 
money  lent  and  repaid  immediately. 

The  money  accounts  should  be  transferred  to  the  Leger  at  least 
every/  month.  When  the  cash  account  is  entered  into  the  Journal, 
it  is  written,  Cash  Dr.  to  Sundries,  for  money  received,  and  Sun- 
dries Dr.  to  Cash,  for  money  paid,  mentioning  all  the  necessary 
particulars. 

The  following  Caj^h  Book  shows  the  manner  in  which  this  book 
js  formed  and  kept.  It  belongs  to  Lxample  2,  of  Double  Entry,  and 
U  the  Cash  Book  referred  to  in  the  Day  or  Waste  Book  of  that 
Example,  The  same  remark  applies  to  the  auxiliary  books  which 
follow  the  Caeh  Book- 

Topost  the  Cash  Book  into  the  Leger, 

Make  Cash  Dr.  to  Sundries  for  the  afriount  received,  and  Cr. 
by  Sundries  for  the  amount  paid  Then  make  each  account  Dr. 
to  (-ash,  for  the  respective  sums  paid,  and  Cr.  by  Cash  for  the 
respective  sums  received.     See  Example  2,  Leger. 


i 


BY  DOUBLE  ENTRY. 


517 


CASH  BOOK.     JANUARY,  1820. 


Jan. 

Dr. 

$ 

c. 

Jan. 

Cr. 

■    $ 

c. 

3 

To  interest  for  discount- 
ing Wm.  Burr's  Bill 

2 

By  charges  on  merchan- 
dise, per   the   Venus 

No.  13. 

€ 

39 

for  Naples, 

23 

14 

Bills  receivable,  No.  11, 

3 

Bills    payable,  No.    13, 

Williams  &  Co. 

1600 

Wm.  Burr, 

1440 

23 

Ship   Fhebus,  received 

7 

Charges  on  mer.  per  the 

for  freight, 

400 

87 

Dolphin,  for  Bilboa, 

32 

5© 

25 

Farm  in  Cambridge,  re- 

12 

Bills    payable,   No.    11, 

ceived  for  produce, 

lao 

35 

George  Myers, 

2222 

26 

Bills  receivable,  No.  12, 

21 

Charges  for  sales,  per  the 

George  Murray, 

1300 

Betsy,  pd.  customs,  &;c. 

439 

24 

31 

Interest,  half  year's  div- 

23 

Bills    payable.   No.    12, 

idend  at  the  bank. 

300 

John  Howe, 

600 

31 

Funded  property,  sold 

25 

Ship  Phebus,  paid  for  re- 

9000 at  811, 

7272 

pairs. 

130 

2S 

31 

Debentures,  received, 

300 

27 

Charges  on  mer.  per  the 
Henry  for  Jamaica, 

51 

50 

11339 

61 

31 

Expeuses  of  House, 

150 

52 

5089)01 

II.    THE  BILL  BOOK. 

The  Bill  Book  is  a  record  of  all  Bills  of  Exchange  receivable 
or  payable.  The  reference  in  the  Day  Book,  is  by  the  initials  B. 
^.  or,  by  B.  R.  for  bills  receivable,  and  B.  P.  for  bills  payable. 

Bills  Receivable  are  tho^e  paid  or  to  be  paid  to  the  merchant. 

Bills  Payable  are  those  drawn  on  the  merchant  or  to  be  paid  by 
him. 

The  particulars  of  each  kind  of  bills  are  entered  in  the  Bill  Book 
under  their  separate  beads  of  B.  R.  or  B.  P. 

The  records  of  the  Bill  Book  are  entered  into  the  Journal,  un- 
<ler  the  heads.  Bills  Receivable  Dr.  to  Sundries,  for  all  bills  ac- 
cepted, and  Sundries  Dr.  to  Bills  Payable,  for  all  bills  to  be  paid, 
with  all  the  necessary  particulars  of  names,  numbers,  &,c. 

To  post  the  Bill  Book  into  the  Leger,  make  bills  receivable  Dr. 
to  sundries  for  their  whole  amount,  and  bills  payable  Cr.  by  sun- 
dries for  their  whole  amount.  Then  make  the  persons  for  whom 
bills  have  heen  accepted,  Dr.  to  bills  payable  for  their  respective 
amounts,  and  each  person  from  whom  bills  received  Cr.  by  bills 
receivable  for  their  respective  amounts. 

The  following  is  a  copy  of  the  Form  of  the  Bill  Bqok  for  Exam- 
ple 2,  of  Double  Entry. 


518 


BOOK  KEEPING 


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BY  DOUBLE  ENTRY.  519 

m.  THE  INVOICE  BOOK. 

An  Invoice  is  an  account  of  merchandise  exported,  wilh  the 
charges  on  the  shipment. 

The  Invoice  Book  contains  every  invoice  of  goods  shipped 
abroad.     It  is  referred  lo  in  the  Day  Book  by  the  initials  I.  B. 

As  an  invoice  accompanies  goods  received  also,  this  book  is 
sometimes  dirstinguished  into  two  general  heads,  Invoice  inward, 
and  Invoice  outward.  If  the  former  be  preserved  on  tile  for  ref- 
erence, it  vviU  be  sufficient  to  enter  into  the  Invoice  Book  merely 
the  invoices  of  goods  exported. 

The  invoice  contains  the  name  of  the  ship,  master,  the  place  of 
destination,  and  of  the  person  to  whom  the  consignment  is  made, 
and  then  the  quantity  of  goods  and  amount  at  prime  cost,  with  the 
shipping  charges,  pn  this  whole  sum  commission  is  charged. 
The  commission,  and  insurance,  if  the  merchandise  be  insured,  is 
then  added  to  the  cost  and  charges,  and  the  drawback,  if  any  is  al- 
lowed at  the  custom  house,  deducted. 

The  record  of  the  Invoice  Book  is  entered  into  the  Journal  in 
the  following  manner. 

The  person,  for  whom  the  the  Invoice  is  sent,  Dr.  to  Sundries, 
viz.  I 

To  merchandise,  for  goods  shipped. 
To  charges,  for  shipping  charges,  &c. 
To  commission,  for  the  commission. 
To  insurance,  for  the  insurance. 

The  following  Invoice,  referred  to  on  page  2  of  Day  Book,  Ex- 
ample 2,  of  Double  Entry,  shows  the  method  of  forming  the  In- 
voice Book,  and  is  a  specimen  of  the  other  invoices  referred  to  in 
the  same  Day  Book,  by  the  initials  I.  B. 

To  post  the  Invoice  Book  into  the  Leger,  make  the  person  to 
whom  the  invoice  is  sent  Dr.  to  sundries,  for  the  amount.  Then 
make  merchandize,  charges,  commission,  &c.  Cr.  for  the  respec- 
tive sums  belonging  to  each. 

The  posting  of  the  Invoice  Book  is  rendered  shorter  and  more 
simple  by  uniting  several  invoices  when  it  can  be  done,  as  on  page 
3,  of  Journal,  Example  2. 


im 


BOOK  KEEPING 


INVOICE  OF  SUGAR, 

Shipped  on  board  the  Venus,  W.Brown  master,  for  Naples,  by 
order  of  George  Parish,  merchant,  on  his  account  and  risk^and  con- 
signed to  him. 

January  2,  1820. 


o.p. 

Cwt.  qr.  Ife.       Cwi.  qr  Ife. 

$ 

c. 

lto3 

No.    1  Gross,   12  0  12  Tare   1   2     3 

2    12  2  16  1  3     0 

3    11   3  24 1   0  25 

-                        ,  „ 

Gross  36  2  24             4  2     0 

Tare     4  2     0 

' 

516 

43 

Neat    32  0  24  at  16  dolls,  per  Cwt. 

Charges. 
Debenture  Entry,               -         *         -          16  00 
Cost  of  hogsheads,               -          -          -            4  76 

Cartage,  wharfage,  bill  of  lading,         -           2  25 

23 
638 

^ 

43 

Commission  on  g538^\3_  ^t  2i  peF  cent. 
Premium  of  Insurance,        -         .         -         - 

13 

8 

46 
40 

560 

29 

Drawback  allowed  at  the  Custom  House, 

120 

(Entered  Journal  page  2.) 

440 

29 

Deducting  the  drawback  from  the  cost  of  the  mer- 
chandise, the  account  in  the  Day  Book  would  be  as 

follows, 

Merchandise,  395  43 

• 

! 
i 

Charges              23  00 
Commission,       13  46 

Insurance,           8  40 

—440  29 

See  page  2,  Day  Book,  Ex.  2,  of  Double  Entry. 

IV.    THE  SALES  BOOK. 

The  object  of  the  Sales  Book  is  to  show  the  net  proceeds  of  any 
goods  received  to  be  sob!  on  commission.  Its  reference  in  the  Day 
Book  is  S.  B. 

Each  account  of  sales  begins  with  the  names  of  the  goods,  ship, 
and  persuti  by  whom  the  consignment  is  made,  and  contains  two 
general  columns  or  pages.  In  the  first  page  are  recorded  the  va- 
rious charges  arising  from  duties,  freight,  landing,  storeage,  com- 
mission, &c.     The  second  contains  the  quantity,  price  and  amount 


BY  DOUBLE  ENTRY. 


521 


«old,  with  the  names  of  the  purchasers  and  the  time  of  payment. 
The  diiference  between  the  amount  of  the  two  columns  or  pages, 
is  the  net  proceeds.  The  account  is  then  to  be  transmitted  to  the 
owner  of  the  goods,  being  copied  from  the  Sales  Book  and  signed 
by  the  agent  to  whom  the  good-*  were  consigned. 

If  the  goods  are  sohl  for  ready  money,  the  account  must  be  en- 
tered into  the  Cash  Book.  • 

In  entering  an  account  of  sales  into  tha  Journal,  the  person,  or 
owner  of  the  goods  is  made  Dr.  for  the  sales.  Then,  Sales  per 
ship is  Dr.  to  Sundries,  viz. 

To  charges  on  merchandise,  for  the  charges  made. 

Interest,  if  any  money  has  been  advanced. 

Commission,  for  the  commission. 

To  the  consignee, for  the  net  proceeds. 

SALES  BOOK. 


A.fcCOtrNT   OF   8   PIPITS    OP   PORT   WINE,   RECEIVED     PER   THE    BETSEY,   FROJV?: 
OPORTO,    AND    SOLD    OJV    ACCOUNT    OF    GEORGE    GREAVES. 


1820 

Dr. 

$ 

c. 

I82U 

Cr. 

v? 

c. 

Jan. 

To  duly  on  1118  galls. 

Jan. 

By    Smith  &  Son,  sold 

21 

at  33i  cents  per  gall. 

2n 

67 

26 

them  payable  iu  Imo. 

Freight,  primage,  &;c. 

30 
403 

3t\ 
00 

3  pipes. 
No.  1,  139  galls. 
2,  141 

Cooperage  at    50    cents 

3,  138 

per  pipe. 

4 

Cartage,  wharfage,  &c. 

16 

418  galls. 

Storeage,  and  Insurance, 

Ullage     •  1    do. 

and  taking  stock, 

10 

Brokerage,  78  cents  per 

417     do.      at 

pipe, 

6 

21 

$154  per  pipe  of  139 

gallons. 

462 

Amount  of  charges. 

43G 

24 

26 

By  Jos.  Lockwood,  sold 

Interest  on  $403  for  60 

him  payable   in  2mo. 

days  at  6  per  cent,  ad- 

5 pipes. 

vanced. 
Commission  at  2|  per  ct. 

4 

03 

No.  1,  140  galls. 
2,  140 

on$1234_y_    .. 

33 

477 

21 

1 

3,  139 

4,  140 

5,  141 

26 

Net  proceeds,  due  this 
day  to  Geo.  Greaves, 

700  galls. 

Oporto, 

756 
1234 

93 
M 

Ullage      3  do. 



697    do.      ai 
$154  per  pipe  of  139 

gallons, 

772 
12:^4 

14 
14 

To  post  the  Sales  Book  into  the  Leger,  make  the  persons  to 

whom  the  consi^-nment  is  made   Dr.  to  Sales  (per  ship )  for 

the  amount.     Then  make  the  consigner,  charges,  commission,  in- 
terest, iic.  Cr.  by  sain?  for  the  snms  !)elonging  to  each  respectively. 


d5J2 


BOOK  KEEPING 


Note.  Besides  the  Auxiliary  Books  already  mentioned,  seve^ 
ral  others  are  occasionally  employed,  which  the  accountant  can 
readily  form  for  himself.     The  names  and  ohject  of  several  follow. 

1.  The  Book  of  Accounts  current,  is  a  record  of  accounts  sent 
to  your  employers. 

2.  The  Book  of  Commissions,  is  a  record  of  Orders  from  corres- 
pondents.      0 

3.  Book  of  Charges  contains  accounts  not  charged  to  any  thing 
else,  as  rent,  wages,  postage,  incidental  expenses. 

4.  Copy  Book  of  Letters,  sent  to  correspondents. 

6.  Book  of  Postage  of  letters,  contains  their  date  and  cost. 

G.  Book  of  Ship  Charges,  contains  the  charges  for  each  ship, 
which  is  to  be  carried  to  the  proper  account  of  the  ship  in  the  Le- 
ge r. 

7.  Book  of  Receipts,  for  all  receipts. 

8.  Memorandum  Book,  for  particulars  to  be  attended  to  after- 
wards. 

EXAMPLE  n.    WASTE  BOOK. 


January  1,  1820. 

Inventory  of  all  my  property  real  and  personal, 
with  a  list  of  th€  balances  in  my  favour  and  against 
me,  taken  this  day. 
Cash  in  hand, 
Funded  property  gl2000  in  the  5  per  ceats,  at  80| 

per  cent, 
Farm  in  Cambridge, 
House  in  the  city, 
Furniture, 

Ship  Phebus,  my  half, 
Merchandise  on  hand, 
Debentures  for  balance  due  at  the  Custom  House, 

Bills  receivable,  the  following  in  hand. 

No.  11.  On  Williams  &  Co.  due  Jan.  14,  g  1600  00 

12.  On  George  Myers,  26,     1300 


Balances  in  my  favour,  viz. 
James  Greaves,  Oporto, 
Cyrus  Coate,  Bordeaux, 
Lemuel  Rogers,  Bilboa, 
George  Parish,  Naples, 


$1700  00 
1660  35 
1175  00 
2200  00 


I  owe  as  follows, 

Bills  payable  for  bills  accepted  by  me. 

No.  1 1.  Drawn  by  George  Myers,  due  Jan.  12,  $2222 

12.  John  Howe,  23,       600 

13.  William  Burr,  30,     1440 


13500 

%60 
4500 
2050 
1200 
9500 
6400 
1300 


2900 


6G35 


5764^ 


4262 


BY  DOUBLE  ENTRY. 
EXAMPLE  II.    WASTE  BOOK. 


5'i3 


Balances  against  me. 

To  Smith  &  Son,  London,                        •  $2150  00 

■  To  Gilson  &  Co.     Do.                                    800  00 

To  Spring  &  Jones,  Jamaica,                        1666  67 

To  George  Black,         Do.                             1175  00 

To  James  Broker,        Do.                            1450  58 

7242 

25 

25 

11504 

January  2d,  1820. 

Shipped  on  board  the  Venus,  W.  Brown  master. 

for  Naples,  sugar  on  the  account  of  George  Par- 

ish, as  per  Invoice  Book,  viz. 

Merchandise,                                           $395  43 

Charges,                                                       23 

Commission,                                                  13  46 

Insurance,                                                       8  40 

440 

29 

2d. 

Received  by  post  a  bill  from  Cyrus  Coate  of  2466 

livres.  at  16f  cents,  as  per  B.  K. 

411 

3d. 

Paid  William  Burr's  bill,  No.  13,  as  per  C.  B. 

1440 

Received  discouot  on  the  above  for  27  days,  at  6 

per  cent   as  per  C  B 

6 

39 

7th. 

Shipped  on  board  the  Dolphin,  for  Bilboa,  goods  on 

account  of  Lemuel  Rogers,  as  per  1.  B. 

Merchandise,                              $2000  00 

Charges,                              ,              32  60 

] 

Commission,                                     60  75 

• 

Insurance,                                        34   15 

2127 

40 

12th. 

Paid  George  Myers  bill,  due  this  day,  No.  11.  as 

per  C.  B. 

2222 

_ 

14lh. 

7 

Received  the  amount  o(  Williams  &  Go's,  bill,  No. 

11,  as  per  C.  B. 

1600 

— 

15th. 

Accepted  a  bill  drawn  by  Gilson  &  Co.  No.  1,  B  P. 

800 

— 

16th. 

Bought  of  Joseph  Lockwood,  sundry  goods  amount- 

ing, as  per  bills  of  parcels,  to 

10000 

87. 

20lh. 

Samuel  Lockwood  has  drawn  on  me,  a  bill,  No.  2, 

as  per  B.  P. 

3600 

BU 


BOOK  KEEPING 


EXAMPLE  II.    WASTE  BGCMv. 


January  21,  1820. 

-— . 

Arrived  the  Betsey  from  Oporto,  with  8  pipes  of 

Port  Wine,  consigned  by  James  Greaves,  to  me, 

to  sell  on  his  accaunt,  S.  B. 

Paid  sundr}'  charges  on  landinc 

439 

24 

22d. 

Received  of  Lemuel  Rogers,  a  bill  of  Exchange, 

No  2,  as  per  B.  R. 

2500 

— 

23d. 

Pa..!  John  Howe's  biil,  No.  12,  C.  B. 

600 

— 

Kticeived  of  G   Seaman,  my  half  snare  for  freight 

on  board  the  ship  Phebiis,  C.  B. 

400 

87 

24th. 

Accepted  a  bill  drawn  by  Smith  k  Son,  of  Lon- 

don, B.  P. 

2150 

25lh. 

Received  of  George  Sabin,  for  produce  of  the  farm 

in  Canibriflge.  C..  B. 

160 

35 

25th. 

Paid  for  repairs  of  the  ship  Phebus,  C.  B. 

130 

25 

26lh. 

Sold  to  Smith  &  Son,  Port  Wine,  S.  B. 

462 

Sold  to  Jo^seph  Lock  wood,  Port  Wine,  S.  B. 

772 

14 

Rt  c^MVf^d  caj-h  orGeor{i»"V Murray's  bill,  No.  12,  C  B. 

1300 

27th. 

Shipped  on  board  the  Henry,  Talbot,  master,  for 

Jamaica,  sundry  goods  for  sundry  persons,  as  per 

I.  B.  viz. 

Spring  &  Jones, 

Merchandise,                                 ^1120  00 

Charges,                                                 12  60 

Commission,                                           35  75 

Insurance,                                             37  25 

1205 

50 

George  Blacfe, 

Merchandise,                                  ^1800  00 

Charges,                                                   15  00 

Coroinission,                                         56  50 

Insurance,                                              62  60 

1934 

1     James  Broker  and  Shipper,  each  half  a  share, 

Merchandise,                                 ^2500  00 

Charges,                                                24  00 

Commission,                                           00  00 

Insurance,                                         105  48 

2719 

48 

BY  DOUBLE  ENTRY. 


EXAMPLE  II.    WASTE  BOOK. 


January  31,  1820. 
Received  a  dividend  at  the  Bank,  half  years  inter- 
est on  gl2000  at  5  per  cent.  C.  B. 


Sold  §9000  of  stock,  at  81  per  cent,  commission 
per  cent.  C  B 


Received  dcbentnres  for  snoods  shipped  this  month, 
Received  cash  for  f»f  henluros  this  month,  C.  B. 


Paid  lor  house  expenses  this  month,  C  B. 


300 


7272 
200 
300 
160 


50 


52 


/ 


BOOJfC  KEEPING 


EXAMPLE  II.    JOURNAL.    JANUARY  1820. 


(1) 


D. 

L. 

P. 

Sundries  Dr,  to  Stock, 

% 

c. 

1 

1 

Cash  in  hand, 

Funded  property  g  12000,  at  80^  per  cent, 

Farm  in  Cambridge, 

House  in  the  city, 

Furniture, 

13600 
9660 
4500 
2060 
1200 

Ship  Phebus,  my  half, 
Merchandise  on  hand, 

9600 
6400, 

2 

Debentures,  balance  due  at  the  Custom  House, 

1300 

2 

Bills  Receivable,  bills  due  me,  amount, 

2900 

2 

James  Greaves,  Oporto, 

1700 

2 

Cyrus  Coate  Bordeaux, 

1660 

36 

2 

Lemuel  Rogers,  Bilboa, 

1176 

j 

2 

George  Parish  Naples, 

2200 

67646 

36 

Stock  Dr.  to  Sundries, 

1 

2 

To  Bills  Payable,  bills  accepted  by  me, 

4262 

2 

Smith  &  Son,  London, 

2160 

3 

Gilson  &  Co,         do. 

800 

2 

Spring  &  Jones,  Jamaica^ 

1666 

67 

2 

George  Black,         do. 

1176 

3 

James  Broker,         do. 

• 

1450 

58 

11504 

25 

Cash  Dr.  to  Sundries, 

For  sums  received  this  month  as  per  C.  B. 

3 

3 

To  Interest,                                                        6  39 

n 

3 

do.                                                         300  00 

306 

3^ 

4 

2 

Bills  Receivable,  No.  11,                   1600  00 

IG 

2 

io                           i«3nn   nn 

4, 

9.^CsV\ 

>3 

1 

Ship  Phebus, 

400  87 

lb 

1 

Farm  in  Cambridge^ 

160 

35 

11 

2 

Debentures, 

300 

31 

1 

Funded  property. 

7272 

11339 

67 

BY  DOUBLE  ENTRY. 


527 


EXAMPLE  11.    JOURNAL.    JANUARY  1820. 


(2) 


Ja. 

P. 
L. 

Sundries  Dr.  to  Cash, 

For  sums  paid  this  month  as  per  C.  B. 

$ 

c. 

2 

1 

Charges  on  merchandise,  viz.  per  Venus,      23  00 

7 

Dninhiri      S^    ^D 

9^ 

Rrf-^pv     4SQ   ^1 

X  1 

97 

Hrnrv        51    50 

^  i 

.546 

24 

3 

2 

Bills  payable,  No.  13,                                   1440  00 

xJ^U 

12 

11,                                    2222  00 

oq 

lo                                            f^an  nn 

^O 

426'!> 

25 

1 

Ship  Phebus, 

130 

25 

31 

3 

House  expenses. 

150 

52 

5089 

^ 

Bills  Receivable  Dr.  to  Sundries, 

For  bills  received  this  month,  as  per  B.  K. 

2 

2 

To  Cyrus  Coate,  No.  1,  due  March  1, 

411 

22 

2 

Lemuel  Rogers,  No.  2,  Feb.  15, 

2500 

2911 

Sundries  Dr.  to  Bills  Payable, 

For  bills  accepted  by  me  this  month,  as  per  B.  P. 

15 

20 

3 

To  Gilson  &  Co.  No.  1,  due  Feb.  1, 
Joseph  Lockwood  No.  2, 3, 

800 
3600 

24 

Smith  &  Son,  No.  3,     March  1, 

2150 

6550 

George  Parish  Dr.  to  Sundries, 

For  amount  of  Invoice  of  Sugar,  per  the  Venus, 

for  Naples,  as  per  I.  B.  and  W.  B. 

2 

1 

1 

2 

To  Merchandise,                                     395  43 
Charges,                                               23  00 
Commission,                                          13  46 
Insurance,                                              8   40 

440 

29 

Samuel  Rogers  Dr.  to  Sundries, 

Amount  of  invoice  per  Dolphin,  for  Bilboa,  as 

per  VV.  B.  p.  2. 

1 

1 

To  Merchandise,^                                2000  00 
Charges,                                                 32  50 

1 

Commission,                                          60  75 

2 

Insurance,                                              34   15 

2127 

40 

BOOK  KEEPING 


EXAMPLE  II.    JOURNAL.    JANUARY  182<i. 


'C>) 


J  a. 
16 


31 


26 


27 


26 


3 


Merchandise  Dr.  to  Jos.  Lockwood, 

To  amount  of  goods  bought  of  him,  as  per  bills  of 
parcels. 


Insurance  Dr.  to  Globe  Insnr.  Co. 

For  amount  of  Insurance,  as  per  I.  B. 
Per  Venus,  for  Naples,  $3  40 

Per  Dolphin,  for  Bilboa,  34   15 

Per  Henry,  for  Jamaica,  206  23 


Debentures  Dr.  to  Merchandise, 


For  drawbacks  received  this  month. 


Sundries  Dr.  to  Sales  per  the  Betsey, 

For  amount  of  8  pipes  of  Port  Wine,  on  account 

of  James  Greaves,  as  per  S.  B. 
To  Smith  &  Son,  3  pipes  at  1  month, 

Joseph  Lockwood,  5  pipe-^  at  2  months, 

Sundries  Dr.  to  Sundries, 

For  amount  of  Invoices  per  the  Henry,  for  Jamai- 
ca, as  per  W.  B.  p.  3. 

Insur. 

Spring  &  Jones, 
George  Black, 
Adventure    to   Ja 

maica,  inCo  with 

J.    Broker,     my 

half, 
James  Broker,  his 

half, 

Sums, 


$     c. 


10000 


87 


247 


200 


Merchan. 
1120  00 

Charg. 
12  50 

Commis. 

35   75 

1800  00 

15  00 

56  50 

2500  00 

24  00 

90  00 

5420  00 

51    50 

U!2   25 

37  25 
62  50 


105  48 


'205  23 


Sales  per  the  Betsey  Dr.  to  Sundries, 

To  charges  on  merchandise,  439  24 

Commission,  33  94 

Interest,  4  03 

James  Greaves,  for  net  proceeds  of 

8  pipes  of  Port  Wine,  as  per  S.  B.  756  93 


462 

772 


120 
193 


1359 


1359 


78 


50 


14 


60 


74 


74 


5858  98 


1234  14 


-\ 


BY  DOUBLE  ENTRY. 


529 


ALPHABETICAL  INDEX  TO  THE  LEGER. 


A. 

Adventure  to  Jam.  3, 


D. 
Debentures, 


G. 

Gilson  &  Co. 
Globe  Ins.  Co. 
Greaves.  James 


B. 

Balance,  3 

Bills  payable,  2 
Bills  receivable,  2 

Black,  Geo.  2 

Broker,  James  3 

E. 


H. 

House, 

House  Expenses, 


C. 

Cash, 

Charges  on  Mer. 
Coate,  Cyrus 
Commission, 


F. 

Farm  in  Camb.  1 
Funded  Property,  1 
Furniture,  1 


I. 

Insurance, 
Interest, 


K. 

N. 


L. 


M. 


Lockwood,  Jos.     3  Merchandise,     1 


O. 


R. 


T. 


Rogers,  Lemuel 


U. 


P. 

Parish,  Geo.         3 
Profit  &  Loss,      3 


Sales  per  Betsey,  3 

Ship  Phebus,  1 

Smith  &  Son,  2 

Spring  &  Jones,  2 

Stock,  1 


W. 


(Cr  Before  attempting  to  balance  the  Leger,  it  must  be  ascertained  whether 
the  Journal  has  been  correctly  formed  from  the  Day  Book  and  Auxiliary  Books, 
and  whether  the  journal  has  been  correctly  posted  into  the  Leger.  In  examin- 
ing the  books  for  this  purpose,  a  point  or  some  mark  should  be  placed  against  the 
several  accounts  found  to  be  correctly  entered  in  the  Journal  and  Leger,  and 
this  pointing  or  marking  continued  through  all  the  accounts.  It  is  then  common 
to  make  a  trial  balance,  on  a  separate  piece  of  paper,  before  forming  the  account 
called  Balance,  in  the  Leger.  If  the  books  can  be  thus  balanced,  the  several 
balances  are  then  placed  under  Balance,  and  the  work  is  finished. 

Note.  The  examples  here  given  are  sufficient  to  illustrate  the  method  of 
Double  Entry.  The  teacher  should  not  suffer  the  pupil  to  pass  over  any  point, 
until  it  is  well  understood.  If  more  examples  should  be  desired,  he  can  direct 
the  learner  to  take  Ex.  2.  of  Single  Entry,  and  form  from  it  the  several  books  in 
Double  Entry.  After  this  has  been  done,  the  pupil  should  form,  for  himself  a 
larger  Day  or  Waste  Book  for  Double  Entry,  and  carsy  Iho  account  throvigh  all 
the  form",  according  to  the  principles  taught  in  this  sv-lpm, 
T  3 


650  BOOK  KEEPING 

(1)  EXAMPLE  II.    LEGER.    JANUARY  1820. 


i).IJ. 
P. 
1 


Dr.     Stock, 
To  Sundries, 
Balance, 


Dr.     Cash, 
To  Stock, 
Sundries, 


Dr.      t  unded 

To  Stock  12000a80i 
Profit  and  Loss, 


Or.     h  arm  in 

To  Stock, 

Profit  and  Loss, 


I  *.  i^ou>4:j  in  ihe 
To  Stock, 


Di\  Furnitur,', 
Po  stock. 


Dr.         JShip 

To  Stock, 
Cash, 
Profit  and  Loss, 


Dr.  Merchandise, 

To  Stock, 

Jos.  Lockwood, 


Dr.  Charges  on 
To  Cash, 


Dr.<'oinxnis!*ion, 
To  Profit  and  Loss, 


$ 

11504 

47049 

5r.5h3 


13500 
ll"339 


'24SrM^ 


9660 

27 


61 


961j: 


4500 
160 


.35 


20501 


1200 


9500 
130 

270 

G900 


6400 
10000 


l<i400 


1   546 


290 


24 


10 


31 


25 


23 


Contra   Cr. 
By  Sundries, 

Profit  and'  Loss, 


|L. 


Ca.  Cj 

By  Sundries, 
Balance, 


l^roperty,  Cr« 

By  Cash  9000  at  80| 

Balance  3000  aSOi 


Cambridge,  Cr 
By  Casli, 
Balance, 


Ciiy,  Cr. 

By  Balance, 
-~-  Cr." 

By  Balance, 


Phf  bus, 
By  Cash, 
Balance,- 


Cr. 


(a.  Cr. 

By  Geo.  Parish, 
Lemuel  Rogers 
Sundries, 
Debenture:?, 
Balance, 


Merchandise, Cr. 
By  George  Parish 

Lemuel  Rogers 

Sundries,   ' 

Sales  per  Betsey, 


Ca.     Cr. 

2  By  George  Parish 
2   Lemuel  Rogers, 
■I   Sundries, 
o    Sales  per  Betsey 


57645 
908 


>8553 


5089 
19750 


24839 


7272 
2415 


9687 


160 
4500 


4660 


2050 
1200 


400 
9500 


9900 


395  43 
2000 
5420 

200  50 
8384  94 


61 


87 


16400  37 

23J 
32I5O 
50 


51 

539 


546 


13 

60 
182 
33  94 


290  40 


BY  DOUBLE 
EXAMPLE  II.    LEGER. 


Eb'TRV. 

JANUAR-Y  1820. 


53i 


Ja 

J. 
P. 

3 

1 

2 

i 

1 
1 

1 

2 

1 

2 

2 

3 

3 

3 

Dr.  Insurance, 
To  Globe  Insuraace 
Company, 

L. 

3 

1 

1 

3 

1 

1 

1 
1 

$ 
247 

c  . 
78 

50 

D. 

Ja 

2 

7 

27 

1 

31 

26 

2 

22 

1 

1 

1 

J. 
P. 

2 
2 
3 

1 

1 

2 

1 

3 

o 
2 

1 
1 
1 

Ca.           Cr. 

By  George  Parish, 
Lemuel  Rogers, 
Sundries, 

L. 

o 

i 
1 

3 
1 

1 

'J 

*    < 

3-1 

205 

40 

15 
23 

247 

2900 
2911 

7S 

L>r.           Bills 

To  Stock, 
Sundries, 

2900 
2911 

Keceiva.ble,  Cr. 

By  Cash, 
Bejance, 

.5811 

5811 

1 

Dr.           Bills 

To  Cash, 
Bakmce, 

4262 
6550 

Payable,     Cr. 

By  Stock, 
Sundries, 

4262 
6550 

10812 

108  li^ 

1 

31 

Dr.  Debentures, 

To  Stock, 

Merchandise, 

1300 

200 

Ca.           Cr. 

By  Cash, 
Balance, 

300 
1200 

1500 

50 

1500 
1700 

50 

35 
40 

50 

) 

Dr.         James 

To  Stock, 

Greaves,    Cr. 
To  Sales  per  B«tsey, 
Balance, 

3 

3 

2 
2 
3 
1 

_ 

1 

3 

756 
943 

03 

171X) 

411 
1149 

1560 

1 

Dy.        Cyms 

To  Stock, 

1560 

Coale,        Cr. 
By  Bills  Receivable, 
Balance, 

jj 
A5 

1 

Dr.      Lenauel 

To  Stock, 
Sundries, 

1175 

2127 

Kogfer^*,      Cr. 

By  Bills  Receivable, 
Balance, 

2500 
802 

3302 
2640 

40 

3302 

2200 
440 

40 
29 

40 

] 

2 

Dr.       George 

To  Stock, 
Sundries, 

] 

2 
3 

3 

Parish,       Cr. 
By  JBalanc*, 

20 

2640 

29 

50 
17 

67 

24 
2G 

i)r,      8milh  &L 
To  Bills  Payable, 
Sales  per  iietsey, 

2150 
462 

!Son,             Cr. 
By  Stock, 
Balance, 

21.50 

462 

2612 

2612 

27 

Dr.     bpnng  it 
To  Sundries, 
Balance, 

1205 
461 

1666 

Jones,         Cr. 
By  Stock, 

IGGQ 

67 

27 

Dr.     George 

To  Sundrie?, 

1934 

hiack,        Cr. 
By  Stock, 
Balance, 

1175 
759 

'   1934 

53:^ 


T. 
p. 


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LP  21A-15m-4.'63 
(U64VlliUJ4''ib 


G<  neral  Library 

■Uaivcnsity  c^  California 

Berkeley 


House, 
Furniture, 
Ship  PLebus, 
Merchandise, 
Bills  Receivable, 
Debentures, 
James  Greaves, 
<  yyrus  Coate, 
Lemuel  Rog'ers, 
Gpcr;^c  Parish, 
Smith  k.  Son, 
Geor-'C  Black, 
Advcii.  to  Jamai. 


1  1 

2050 

1 

1 

1 

1200 

1 

] 

9500 

1 

8384 

94 

i> 

2911 

2 

1200  50 

2 

043  07 

2 

1149  35 

i 

802 

4C 

i 

2 

2640 

29 

i 

9 

462 

1 

2 

751) 

1  ! 

3 
1 

135:-|7.:|! 
60027 1  81 

Jos.  Lockwood, 
Globe  Insur.  Co. 

Stock, 


1450 

.l:?59 
0000 


il232 


800 


306 

4 

310 


247 


150 


27 
160 
270 
290 
310 

•058 


6550 
>  461 
\     90 

5628 

247 

47049 


58 


21 


42 


78 


52 


60027  89 


f  y  cl5796' 


882173 

THE  UNIVERSITY  OF  CALIFORNIA  UBRARY 


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